ELEMENTS  OF  HYDRAULICS 


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I 


ELEMENTS 


OF 


HYDRAULICS 


BY 
S.  E.  SLOCUM,  B.  E.,  PH.  D., 

PROFESSOR   OF  APPLIED   MATHEMATICS   IN   THE 
UNIVERSITY   OF   CINCINNATI 


FIRST  EDITION 


McGRAW-HILL  BOOK  COMPANY,  INC. 
239  WEST  39TH  STREET,  NEW  YORK 

6  BOUVERIE  STREET,  LONDON,  E.  C. 

1915 


COPYRIGHT,  1915,  BY  THE 
MCGRAW-HILL  BOOK  COMPANY,  INC. 


THE. MAPLE. PRESS. YORK. PA 


PREFACE 

THE  remarkable  impetus  recently  given  to  hydraulic  develop- 
ment in  this  country  has  caused  the  whole  subject  to  assume  a  new 
aspect.  Not  only  is  this  apparent  in  new  and  improved  construc- 
tion details,  but  in  the  scientific  study  which  is  beginning  to  be 
given  a  subject  which  seemed  to  have  crystallized  into  a  set  of 
empirical  formulas. 

Such  comprehensive  plans  as  those  recently  undertaken  by  the 
State  of  New  York  and  the  Dominion  of  Canada  for  the  system- 
atic development  of  all  their  available  water  power,  indicates  the 
extent  of  the  field  now  opening  to  the  hydraulic  engineer.  The 
extent  and  cheapness  of  the  natural  power  obtained  not  only  from 
the  development  of  existing  streams  but  also  from  the  artificial 
pondage  of  storm  water  is  sufficient  to  convince  even  the  most 
casual  observer  that  no  phase  of  conservation  will  have  a  more 
immediate  effect  on  our  industrial  development  or  be  more  far 
reaching  in  its  consequences. 

The  present  text  is  intended  to  be  a  modern  presentation  of  the 
fundamental  principles  of  hydraulics,  with  applications  to  recent 
important  works  such  as  the  Catskill  aqueduct,  the  New  York 
State  barge  canal,  and  the  power  plants  at  Niagara  Falls  and 
Keokuk.  Although  the  text  stops  short  of  turbine  design,  the 
recent  work  of  Zowski  and  of  Baashuus  is  so  presented  as  to  en- 
able the  young  engineer  to  make  an  intelligent  choice  of  the  type 
of  development  and  selection  of  runner. 

In  order  to  make  the  book  of  practical  working  value,  a  col- 
lection of  typical  modern  problems  is  added  at  the  end  of  each 
section,  and  a  set  of  the  most  useful  hydraulic  data  has  been 
compiled  and  is  tabulated  at  the  end  of  the  volume. 

S.  E.  SLOCUM. 

CINCINNATI,  OHIO, 
January,  1915. 


CONTENTS 

PAGE 

PREFACE    v 

SECTION  I 
HYDROSTATICS 

ART.  1.  PROPERTIES  OF  A  PERFECT  FLUID      1 

Definition — Distinction  between  liquid  and  gas — Elasticity 
of  water — Fluid  pressure  normal  to  surface — Vicosity — 
Density  of  water — Specific  weight. 

2.  HYDROSTATIC  PRESSURE 3 

Equal  transmission  of  pressure — Pressure  proportional  to 
area — Hydraulic  press — Frictional  resistance  of  packing — 
Efficiency  of  hydraulic  press. 

3.  SIMPLE  PRESSURE  MACHINES .       8 

Hydraulic  intensifier — Hydraulic  accumulator — Hydraulic 
jack — Hydraulic  crane — Hydraulic  elevator. 

4.  PRESSURE  ON  SUBMERGED  SURFACES ^    .    .      12 

Change  of  pressure  with  depth — Pressure  on  submerged 
area — Center  of  pressure — General  formula  for  center  of 
pressure — Application. 

5.  EQUILIBRIUM  OF  Two  FLUIDS  IN  CONTACT 16 

Head  inversely  proportional  to  specific  weight — Water 
barometer — Mercury  barometer — Piezometer — Mercury 
pressure  gage. 

6.  EQUILIBRIUM  OF  FLOATING  BODIES 20 

Buoyancy — Floating  equilibrium — Theorem  of  Archi- 
medes— Physical  definition  of  specific  weight — Determina- 
tion of  specific  weight  by  experiment — Application  to 
alloy — Zero  buoyancy. 

7.  METACENTER '    • 23 

Stability  of  floating  body — Metacenter — Coordinates  of 
metacenter — Metacentric  height — Period  of  oscillation — 
Rolling  and  pitching. 
APPLICATIONS .       28 

SECTION  II 
HYDROKINETICS 

ART.  8.  FLOW  OF  WATER  FROM  RESERVOIRS  AND  TANKS 44 

Stream  line — Ideal  velocity  head — Torricelli's  theorem- 
Actual  velocity  of  flow — Contraction  coefficient — Efflux 
vii 


viii  CONTENTS 

ART.  PAGE 

coefficient — Effective  head — Discharge  from  large  rec- 
tangular orifice— Discharge  of  a  rectangular  notch  weir. 

9.  DISCHARGE  THROUGH  SHARP-EDGED  ORIFICE 48 

Contraction  of  jet — Complete  contraction — Partial  con- 
traction— Velocity  of  approach. 

10.  RECTANGULAR  NOTCH  WEIRS 51 

Contracted  weir — Suppressed  weir — Empirical  weir  formu- 
las— Rational  weir  formula. 

11.  STANDARD  WEIR  MEASUREMENTS •  .     53 

Construction  of  weir — Hook  gage — Location  of  hook  gage 
— Proportioning  of  weirs. 

'12.  TIME  REQUIRED  FOR  FILLING  AND  EMPTYING  TANKS  .....     56 
Change  in  level  under  constant  head — Varying  head — 
Canal  lock — Rise  and  fall  in  connected  tanks — Mariotte's 
flask. 

13.  FLOW  THROUGH  SHORT  TUBES  AND  NOZZLES     ......'..        60 

Standard  mouthpiece — Stream  line  mouthpiece — Borda 
mouthpiece — Diverging  conical  mouthpiece — Venturi  ad- 
jutage— Converging  conical  mouthpiece — Fire  nozzles. 

14.  KINETIC  PRESSURE  IN  A  FLOWING  LIQUID    .......    1    ..     64 

Kinetic  pressure — Bernoulli's  theorem — Kinetic  pressure 
head — Application  to  standard  mouthpiece. 

15.  VENTURI  METER   .......    '. 67 

Principle  of  operation — Formula  for  flow — Commercial 
meter — Catskill  aqueduct  meter — Rate  of  flow  controller. 

16.  FLOW  OF  WATER  IN  PIPES 71 

Critical  velocity — Viscosity  coefficient — Parallel  (non- 
sinuous)  flow — Average  velocity  of  flow  in  small  pipes  — 
Loss  of  head  in  small  pipes — Ordinary  pipe  flow. 

17.  PRACTICAL  FORMULAS  FOR  Loss  OF  HEAD  IN  PIPE  FLOW  ...     76 

Effective  and  lost  head — Friction  loss — Bends  and  elbows 
— Enlargement  of  section — Contraction  of  section — Gate 
valve  in  circular  pipe — Cock  in  circular  pipe — Throttle 
valve  in  circular  pipe — Summary  of  losses — Application. 

18.  HYDRAULIC  GRADIENT  .    .......;.. 82 

Kinetic  pressure  head — Slope  of  hydraulic  gradient — 
Peaks  above  hydraulic  gradient. 

19.  HYDRAULIC  RADIUS  ........;.'.... 84 

Definition — Chezy's  formula. 

20.  DIVIDED  FLOW 86 

Compound  pipes — Branching  pipes. 

21.  FIRE  STREAMS ....;....     89 

Freeman's  experiments — Formulas  for  discharge — Height 
of  effective  fire  stream. 

22.  EXPERIMENTS  ON  THE  FLOW  OF  WATER '90 

Verification  of  theory  by  experiment — Method  of  conduct- 
ing experiments — Effect  of  sudden  contraction  or  enlarge- 


CONTENTS  ix 

ART.  PAGE 

ment — Disturbance  produced  by  obstacle  in  current — 
Stream  line  motion  in  thin  film — Cylinder  and  flat  plate — 
Velocity  and  pressure. 

23.  MODERN  SIPHONS 95 

Principle  of  operation — Siphon  spillway — Siphon  lock. 

24.  FLOW  IN  OPEN  CHANNELS 100 

Open  and  closed  conduits — Steady  uniform  flow — Kutter's 
formula — Limitations  to  Kutter's  formula — Bazin's  for- 
mula— Kutter's  simplified  formula. 

25.  CHANNEL  CROSS  SECTION 103 

Condition  for  maximum  discharge— Maximum  hydraulic 
efficiency — Regular  circumscribed  polygon — Properties  of 
circular  and  oval  sections. 

26.  FLOW  IN  NATURAL  CHANNELS 106 

Stream  gaging — Current  meter  measurements — Float 
measurements — Variation  of  velocity  with  depth — Calcu- 
lation of  discharge. 

27.  THE  PITOT  TUBE 109 

Description  of  instrument — Darcy's  modification  of  Pitot's 
tube — Pitometer — Pitot  recorders — Theory  of  the  impact 
tube — Construction  and  calibration  of  Pitot  tubes — 
DuBuat's  paradox. 

28.  NON-UNIFORM  FLOW — BACKWATER 118 

Energy  equation  for  stream  of  variable  depth — Differential 
equation  of  surface  profile — Backwater  curve  for  broad 
shallow  stream — Integration  of  backwater  function — 
Values  of  integral — Condition  for  singularities  in  back- 
water curve — Numerical  application. 
APPLICATIONS 126 

SECTION  III 
HYBRODYNAMICS 

ART.  29.  PRESSURE  OF  JET  AGAINST  STATIONARY  DEFLECTING  SURFACE  144 
Normal  impact  on  plane  surface — Relation  of  static  to 
dynamic  pressure — Oblique  impact  on  plane  surface — 
Axial  impact  on  surface  of  revolution — Complete  reversal 
of  jet — Deflection  of  jet — Dynamic  pressure  in  pipe  bends 
and  elbows. 

30.  PRESSURE  EXERTED  BY  JET  ON  MOVING  VANE 149 

Relative  velocity  of  jet  and  vane — Work  done  on  moving 
vane— Speed  at  which  work  becomes  a  maximum-Maximum 
efficiency  for  single  vane — Maximum  efficiency  for  con- 
tinuous succession  of  vanes — Impulse  wheel,  direction  of 
vanes  at  entrance  and  exit — Work  absorbed  by  impulse 
wheel. 

31.  REACTION  OF  A  JET  .  154 


X  CONTENTS 

ART.  PAGE 

Effect  of  issuing  jet  on  equilibrium  of  tank — Energy  of  flow 
absorbed  by  work  on  tank — Principle  of  reaction  turbine 
— Barker's  mill. 

32.  TYPES  OF  HYDRAULIC  MOTORS 156 

Current  wheels — Impulse  wheels — Reaction  turbines — 
Classification  of  reaction  turbines — Classification  of  hy- 
draulic motors. 

33.  CURRENT  AND  GRAVITY  WHEELS 158 

Current  wheels — Undershot  wheels — Poncelet  wheels — 
Breast  wheels — Overshot  wheels. 

34.  IMPULSE  WHEELS  AND  TURBINES 159 

Pelton  wheel — Efficiency  of  Pelton  wheel — Girard  impulse 
turbine — Power  and  efficiency  of  Girard  turbine. 

35.  REACTION  TURBINES 169 

Historical  development — Mixed  flow  or  American  type — 
Use  of  draft  tube — Recent  practice  in  turbine  setting. 

36.  CHARACTERISTICS  OF  IMPULSE  WHEELS  AND  REACTION  TUR- 
BINES     .    ..;.,..    .    .    ;    .    .    .-.•.->."..=.'. 

Selection  of  type — Action  and  reaction  wheels — Speed  181 
criterion — Capacity  criterion — Characteristic  speed — Spe- 
cific discharge — Specific  power — Specific  speed — Relation 
between  characteristic  speed  and  specific  speed — Classi- 
fication of  reaction  turbines — Classification  of  impulse 
wheels — Numerical  applications — Normal  operating  range 
— Selection  of  stock  runner. 

37.  POWER  TRANSMITTED  THROUGH  PIPE  LINE  AND  NOZZLE   .    .    .    195 

Effective  head  at  nozzle — Velocity  of  flow  for  maximum 
power — Maximum  efficiency — Diameter  of  nozzle  for 
maximum  output  of  power — Graphical  relation  between 
power  and  efficiency — Method  of  determining  nozzle 
diameter. 

38.  EFFECT  OF  TRANSLATION  AND  ROTATION 198 

Equilibrium  under  horizontal  linear  acceleration — Equi- 
librium under  vertical  linear  acceleration — Free  surface  of 
liquid  in  rotation — Depression  of  cup  below  original  level 
in  open  vessel — Depression  of  cup  below  original  level 
in  closed  vessel — Practical  applications. 

39.  WATER  HAMMER  IN  PIPES 202 

Increase  in  pressure  due  to  suddenly  checking  flow — Bulk 
modulus  of  elasticity  of  water — Pressure  waves  in  pipe 
produced  by  suddenly  checking  flow — Velocity  of  com- 
pression wave — Period  of  compression  wave — Increase  in 
pressure  produced  by  instantaneous  stoppage  of  flow — 
Joukovsky's  experiments — Gibson's  experiments. 

40.  HYDRAULIC  RAM .;..........   205 

Principle  of  operation — Efficiency  of  ram. 


CONTENTS  xi 

ART.  PAGE 

41.  DISPLACEMENT  PTJMPS 207 

Suction  pump — Maximum  suction  lift — Force  pump — 
Stress  in  pump  rod — Direct  driven  steam  pump — Calcula- 
tion of  pump  sizes — Power  required  for  operation — Diam- 
eter of  pump  cylinder. 

42.  CENTRIFUGAL  PUMPS 215 

Historical  development — Principle  of  operation — Impeller 
types — Conversion  of  kinetic  energy  into  pressure — Volute 
casing — Vortex  chamber — Diffusion  vanes — Stage  pumps. 

43.  PRESSURE  DEVELOPED  IN  CENTRIFUGAL  PUMP 224 

Pressure  developed  in  impeller — Pressure  developed  in 
diffusor — General  formula  for  pressure  head  developed. 

44.  CENTRIFUGAL  PUMP  CHARACTERISTICS 227 

Effect  of  impeller  design  on  operation — Rising  and  droop- 
ing characteristics — Head  developed  by  pump — Effect  of 
throttling  the  discharge — Numerical  illustration. 

45.  EFFICIENCY  AND  DESIGN  OF  CENTRIFUGAL  PUMPS 234 

Essential  features  of  design — Hydraulic  and  commercial 
efficiency. 

46.  CENTRIFUGAL  PUMP  APPLICATIONS 236 

Floating   dry   docks — Deep   wells — Mine   drainage — Fire 
pumps — Hydraulic  dredging — Hydraulic  mining. 
APPLICATIONS 242 

SECTION  IV 
HYDRAULIC  DATA  AND  TABLES 

TABLE    1.  Properties  of  water 260 

2.  Head  and  pressure  equivalents  .    .............  261 

3.  Discharge  equivalents . '.    .    .    .  262 

4.  Weights  and  measures     .;....., 263 

5.  Specific  weights  of  various  substances .    . 264 

6.  Dimensions  of  steam,  gas,  and  water  pipe  .    .    ...    .    .    .  265 

7.  Capacity  of  reciprocating  pumps 266 

8.  Circumferences  and  areas  of  circles 268 

9.  Efflux  coefficients  for  circular  orifice     ....    .    .    .    .    .    .  273 

10.  Efflux  coefficients  for  square  orifice .    .    .    .    .  272 

11.  Fire  streams 275 

12.  Coefficients  of  pipe  friction 277 

13.  Friction  head  in  pipes 278 

14.  Bazin's  values  of  Chezy's  coefficient 282 

15.  Kutter's  values  of  Chezy's  coefficient 283 

16.  Discharge  coefficients  for  rectangular  notch  weirs 285 

17.  Discharge  per  inch  of  length  over  rectangular  notch  weirs  .  286 

18.  Discharge  per  foot  of  length  over  rectangular  notch  weirs  .  287 

19.  Principles  of  mechanics 288 

20.  Discharge  per  foot  of  length  over  suppressed  weirs    ....  290 
INDEX  .  291 


ELEMENTS  OF  HYDRAULICS 

SECTION  I 

HYDROSTATICS 

1.  PROPERTIES  OF  A  PERFECT  FLUID 

Definition. — -A  fluid  is  defined,  in  general,  as  a  substance  which 
offers  no  resistance  to  change  in  form  provided  this  deformation 
is  not  accompanied  by  change  in  volume.  The  fundamental 
property  of  a  fluid  is  the  perfect  mobility  of  all  its  parts. 

Most  of  the  applications  of  the  mechanics  of  fluids  relate  to 
water,  and  this  domain  of  mechanics  is  therefore  usually  called 
hydraulics  or  hydromechanics.  It  is  convenient  to  subdivide 
the  subject  into  hydrostatics,  relating  to  water  at  rest;  hydro- 
kinetics,  relating  to  water  in  motion;  and  hydrodynamics,  relating 
to  the  inertia  forces  exerted  by  fluids  in  motion,  and  the  energy 
available  from  them. 

Distinction  between  Liquid  and  Gas. — A  liquid  such  as  water 
has  a  certain  degree  of  cohesion,  causing  it  to  form  in  drops, 
whereas  a  gaseous  fluid  tends  to  expand  indefinitely.  A  gas  is 
therefore  only  in  equilibrium  when  it  is  entirely  enclosed.  In 
considering  elastic  fluids  such  as  gas  and  steam,  it  is  always 
necessary  to  take  account  of  the  relation  between  volume  and 
pressure.  For  a  constant  pressure  the  volume  also  changes 
greatly  with  the  temperature.  For  this  reason  the  mechanics  of 
gases  is  concerned  chiefly  with  heat  phenomena,  and  forms  a 
separate  field  called  thermodynamics,  lying  outside,  and  sup- 
plementary to,  the  domain  of  ordinary  mechanics. 

Elasticity  of  Water. — Water,  like  other  fluids,  is  elastic,  and 
under  heavy  pressure  its  volume  is  slightly  diminished.  How- 
ever, since  a  pressure  of  one  atmosphere,  or  14.7  Ib.  per  square 

inch,  exerted  on  all  sides  only  decreases  its  volume  about  20  QQQ* 

and  a  pressure  of  3000  Ib.  per  square  inch,  only  1  per  cent.,  it  is 
customary  in  all  practical  calculations,  and  sufficiently  accurate 

1 


2       £,  i  t/j  >  t  v .-ELEMENTS  OF  HYDRAULICS 

.  for  ordinary  purposes,  to  assume  that  it  is  incompressible.  The 
volume  of  an  ideal  liquid  is  therefore  assumed  to  remain  constant. 

Fluid  Pressure  Normal  to  Surface. — Since  a  perfect  fluid  is 
one  which  offers  no  resistance  to  change  in  form,  it  follows  that 
the  pressure  on  any  element  of  surface  of  the  fluid  is  everywhere 
normal  to  the  surface:  To  prove  this  proposition,  consider  any 
small  portion  of  a  fluid  at  rest,  say  a  small  cube.  Since  this  cube 
is  assumed  to  be  at  rest,  the  forces  acting  on  it  must  be  in  equi- 
librium. In  the  case  of  a  fluid,  however,  the  general  conditions 
of  equilibrium  are  necessary  but  not  sufficient,  since  they  take  no 
account  of  the  fact  that  the  fluid  offers  no  resistance  to  change 
in  form.  Suppose,  therefore,  that  the  small  cube  under  con- 
sideration undergoes  a  change  in  form  and  position  without  any 
change  in  volume.  Since  the  fluid  offers  no  resistance  to  this 
deformation,  the  total  work  done  on  the  elementary  cube  in 
producing  the  given  change  must  be  zero. 

In  particular,  suppose  that  the  cube  is  separated  into  two  parts 
by  a  plane  section,  and  that  the  deformation  consists  in  sliding 
one  of  these  parts  on  the  other,  or  a  shear  as  it  is  called.  Then 
in  addition  to  the  forces  acting  on  the  outside  of  each  part,  it  is 
necessary  to  consider  those  acting  across  the  plane  section.  But 
the  total  work  done  on  each  part  separately  must  be  zero  inde- 
pendently of  the  other  part,  and  also  the  total  work  done  on  the 
entire  cube  must  be  zero.  Therefore,  by  subtraction,  the  work 
done  by  the  forces  acting  across  the  plane  section  must  be  zero. 
But  when  any  force  is  displaced  it  does  work  equal  in  amount  to 
the  product  of  this  displacement  by  the  component  of  the  force 
in  the  direction  of  the  displacement.  Therefore  if  the  work  done 
by  the  force  acting  across  the  plane  section  of  the  cube  is  zero, 
this  force  can  have  no  component  in  the  plane  of  the  section,  and 
must  therefore  be  normal,  i.e.,  perpendicular,  to  the  section. 

Viscosity. — This  absence  of  shear  is  only  rigorously  true  for 
an  ideal  fluid.  For  water  there  is  a  certain  amount  of  shear  due 
to  internal  friction,  or  viscosity,  but  it  is  so  small  as  to  be  practi- 
cally negligible.  The  greater  the  shear  the  more  viscous  the 
fluid  is  said  to  be,  and  its  amount  may  be  taken  as  a  measure  of 
viscosity.  It  is  found  by  experiment  that  the  internal  friction 
depends  on  the  difference  in  velocity  between  adjacent  particles, 
and  for  a  given  difference  in  velocity,  on  the  nature  of  the  fluid. 
The  viscosity  of  fluids  is  therefore  of  great  importance  in  consider- 
ing their  motion,  but  does  not  affect  their  static  equilibrium. 


HYDROSTATICS  3 

For  any  fluid  at  rest,  the  pressure  is  always  normal  to  any  element 
of  surface. 

Density  of  Water.  —  In  hydraulic  calculations  the  unit  of  weight 
may  be  taken  as  the  weight  of  a  cubic  foot  of  water  at  its  tempera- 
ture of  greatest  density,  namely,  39°  F.  or  4°  C.  It  is  found  by 
accurate  measurement  that  a  cubic  foot  of  water  at  39°  F.  weighs 
62.42  Ib.  This  constant  will  be  denoted  in  what  follows  by  the 
Greek  letter  7.  In  all  numerical  calculations  it  must  be  remem- 
bered therefore  that 

7  =  62.4  Ib.  per  cubic  foot.  (i) 

The  density  and  volume  of  water  at  various  temperatures  are 
given  in  Table  1. 

Specific  Weight.  —  The  weights  of  all  substances,  whether 
liquids  or  solids,  may  be  expressed  in  terms  of  the  weight  of  an 
equal  volume  of  water.  This  ratio  of  the  weight  of  a  given 
volume  of  any  substance  to  that  of  an  equal  volume  of  water  is 
called  the  specific  weight  of  the  substance,  and  will  be  denoted  in 
what  follows  by  s.  For  instance,  a  cubic  foot  of  mercury  weighs 
848.7  Ib.,  and  its  specific  weight  is  therefore 

848  7 


Its  exact  value  at  0°  C.  is  s  =  13.596,  as  may  be  found  in  Table  1. 
The  weight  of  1  cu.  ft.  of  any  substance  in  terms  of  its  specific 
weight  is  then  given  by  the  relation 

Weight  =  75  =  62.45  Ib.  per  cubic  foot.  (2) 


2.  HYDROSTATIC  PRESSURE 

Equal  Transmission  of  Pressure. — The  fundamental  principle 
of  hydrostatics  is  that  when  a  fluid  at  rest  has  pressure  applied  to 
any  portion  of  its  surface,  this  pressure  is  transmitted  equally  to 
all  parts  of  the  fluid. 

To  prove  this  principle  consider  any  portion  of  the  fluid  limited 
by  a  bounding  surface  of  any  form,  and  suppose  that  a  small 
cylindrical  portion  is  forced  in  at  one  point  and  out  at  another, 
the  rest  of  the  boundary  remaining  unchanged  (Fig.  1).  Then 
if  AA  denotes  the  cross-sectional  area  of  one  cylinder  and  An 
its  height,  its  volume  is  AA-An.  Similarly,  the  volume  of  the 


4  ELEMENTS  OF  HYDRAULICS 

other  cylinder  is  AA'-An',  and,  since  the  fluid  is  assumed  to  be 
incompressible, 

AA-An  =  AA'-An'. 

Now  let  p  denote  the  unit  pressure  on  the  end  of  the  first  cylinder, 

i.e.,  the  intensity  of  pressure,  or  its  amount  in  pounds  per  square 

inch.  Then  the  total  pressure 
normal  to  the  end  is  pAA,  and  the 
work  done  by  this  force  in  moving 
the  distance  An  is  pAA-An.  Sim- 
ilarly, the  work  done  on  the  other 
cylinder  is  p'AA'-An'.  Also,  if  7 
denotes  the  heaviness  of  the  fluid 
per  unit  volume,  the  work  done  by 
gravity  in  moving  this  weight 
7AA-An  through  the  distance  h, 
F  -  where  h  denotes  the  difference  in 

level  between  the  two  elements 

considered,  is  yAA-An-h.     Therefore,  equating  the  work  done  on 

the  fluid  to  that  done  by  it,  we  have 


n  +  7AA-An-/i  =  p'-AA'-An'. 
Since  AA-An  =  AA'-An',  this  reduces  to 

p'  =  p  +  7h.  (3) 

If  h  —  0,  then  p'  =  p.  Therefore  the  pressure  at  any  point  in 
a  perfect  fluid  is  the  same  in  every  direction.  Also  the  pressure 
at  the  same  level  is  everywhere  the  same. 

Moreover,  if  the  intensity  of  pressure  p  at  any  point  is  increased 
by  an  amount  w,  so  that  it  becomes  p  +  w,  then  by  Eq.  (3) 
the  intensity  of  pressure  at  any  other  point  at  a  difference  of  level 
h  becomes 


P"  =  (p  +  w 
But  since  p'  =  p  +  yh,  we  have  by  subtraction, 

Ptf  =  p'  +  w, 

that  is,  the  intensity  of  pressure  at  any  other  point  is  increased 
by  the  same  amount  w.  A  pressure  applied  at  any  point  is  there- 
fore transmitted  equally  to  all  parts  of  the  fluid. 


HYDROSTATICS 


For  fluids  such  as  gas  and  steam  the  term  yh  is  negligible,  and 
consequently  for  such  fluids  the  intensity  of  pressure  may  be 
assumed  to  be  everywhere  the  same. 

Pressure  Proportional  to  Area.  —  To  illustrate  the  application 
of  this  principle  consider  a  closed  vessel  or  tank,  filled  with  water, 
having  two  cylindrical  openings  at  the  same  level,  closed  by 
movable  pistons  (Fig.  2).  If  a  load  P  is  applied  to  one  piston, 
then  in  accordance  with  the  result  just  proved  there  is  an  increase 
of  pressure  throughout  the  vessel  of  amount 


p  = 


where  A  denotes  the  area  of  the  piston.     The  force  P'  exerted 
on  the  second  piston  of  area  A'  is  therefore 

PA ' 
P'  =  PA'  =  ^f, 


whence 


p; 
p 


FIG.  2. 


FIG.  3. 


The  two  forces  considered  are  therefore  in  the  same  ratio  as  their 
respective  areas.  This  relation  remains  true  whatever  shape  the 
ends  of  the  pistons  may  have,  the  areas  A  and  A'  in  any  case 
being  the  cross-sectional  areas  of  the  openings.  For  instance, 
an  enlargement  of  the  end  of  the  piston,  such  as  shown  in  Fig.  3, 
has  no  effect  on  the  force  transmitted,  since  the  upward  and 

'' 


downward  pressures  on  the  ring  of  area 
ird2  as  the  effective  area. 


•. 
cancel,  leaving 


6 


ELEMENTS  OF  HYDRAULICS 


Hydraulic  Press. — An  important  practical  application  of  the 
law  of  hydrostatic  pressure  is  found  in  the  hydraulic  press.  In 
its  essential  features  this  consists  of  two  cylinders,  one  large  and 
one  small,  each  fitted  with  a  piston  or  plunger,  and  connected  by 
a  pipe  through  which  water  can  pass  from  one  cylinder  to  the 
other  (Fig.  4).  Let  p  denote  the  intensity  of  pressure  within  the 
fluid,  D,  d  the  diameters  of  the  two  plungers,  P  the  load  applied 


FIG.  4. 

to  one  and  W  the  load  supported  by  the  other,  as  indicated  in 
the  figure.     Then 


and  consequently 


W      D 


,  s 


If  the  small  plunger  moves  inward  a  distance  hy  the  large  one 
will  be  forced  out  a  distance  H  such  that  each  will  displace  the 
same  volume,  or 


whence 


4 


Neglecting  friction,  the  work  done  by  the  force  P  in  moving 
the  distance  h  is  then 


HYDROSTATICS 


and  is  therefore  equal  to  the  work  done  in  raising  W  the  dis- 
tance H. 

Frictional  Resistance  of  Packing. — Usually,  however,  there  is 
considerable  frictional  resistance  to  be  overcome,  since  for 
high  pressures,  common  in  hydraulic  presses,  heavy  packing  is 
necessary  to  prevent  leakage.  One  form  of  packing  ex- 
tensively used  is  the  U  leather  packing  shown  in  Fig.  5.  In 
this  form  of  packing  the  water 
leaking  past  the  plunger,  or  ram 
as  it  is  often  called,  enters  the 
leather  cup  pressing  one  side 
against  the  cylinder  and  the 
other  against  the  ram,  the  pres- 
sure preventing  leakage  being 
proportional  to  the  pressure  of 
the  water. 

To  take  into  account  the  fric- 
tional resistance  in  this  case,  let 
ju  denote  the  coefficient  of  fric- 
tion between  leather  and  ram, 
C,  c  the  depths  of  the  packing 
on  the  large  and  small  rams 
(Fig.  4),  and  p  the  intensity 
of  water  pressure.  Then  the 
area  of  leather  in  contact  with  the  large  ram  is  irDC  and  its 
frictional  resistance  is  therefore  irDCp^.  Similarly,  the  frictional 
resistance  for  the  small  ram  is  irdcpfjL.  Consequently 


and 

whence 

W      TV/1  ~4MD\ 

(5) 


Efficiency  of  Hydraulic    Press. — The  efficiency  of    any  ap- 
paratus or  machine  for  transforming  energy  is  defined  as 

.  _          Useful  work  or  effort 

Efficiency  -  Total  available  work  or  effort' 


8 


ELEMENTS  OF  HYDRAULICS 


and  is  therefore  always  less  than  unity.  In  the  present  case 
if  there  were  no  frictional  resistance  the  relation  between  W 
and  P  would  be  given  by  Eq.  (4).  The  efficiency  for  this  type 
of  press  is  therefore  the  ratio  of  the  two  equations  (5)  and  (4),  or 


Efficiency  = 


-4M) 


(6) 


3.  SIMPLE  PRESSURE  MACHINES 


Hydraulic  Intensifier.- 
in  Art.  2   there   are    a 


High  Pressure 
Outlet 


-Besides  the  hydraulic  press  described 
number  of    simple   pressure   machines 
based  on  the  principle  of  equal 
distribution  of  pressure  through- 
".-«    out   a   liquid.     Four   types   are 
here   illustrated  and   described, 
as  well  as  their  combination  in  a 
hydraulic  installation. 

When  a  hydraulic  machine 
such  as  a  punch  or  riveter  is 
finishing  the  operation,  it  is  re- 
quired to  exert  a  much  greater 
force  than  at  the  beginning  of 
the  stroke.  To  provide  this  in- 
crease in  pressure,  an  intensifier 
is  used  (Fig.  6).  This  consists 
of  several  cylinders  telescoped 
one  inside  another.  Thus  in 
Fig.  6,  which  shows  a  simple 
form  of  intensifier,  the  largest 
cylinder  A  is  fitted  with  a  ram 
B.  This  ram  is  hollowed  out  to 
form  another  cylinder  C,  fitted 
with  a  smaller  ram  D,  which  is 
fixed  to  the  yoke  at  the  top. 

In  operation,  water  at  the 
ordinary  pump  pressure  enters 

the  cylinder  A  through  the  intake,  thereby  forcing  the  ram  B  up- 
ward. This  has  the  effect  of  forcing  the  ram  D  into  the  cylinder 
C,  and  the  water  in  C  is  thereby  forced  out  through  D,  which  is 
hollow,  at  an  increased  pressure.  Let  p\  denote  the  pressure 


Low  Pressure 
Intake 


FlG.   6. 


HYDROSTATICS 


9 


of  the  feed  water,  p2  the  intensified  pressure  in  C  and  di,  d2  the 
diameters  of  the  cylinders  B  and  C  respectively,  as  indicated 
in  the  figure.  Then 


Pi  = 


whence 


en 


I       I 


I       I 


The  intensity  of  pressure  in  the  cylinders  is  therefore  inversely 

proportional   to  the  areas  of  the 

n  n 

rams. 

When  a  greater  .intensification  of 
pressure  is  required  a  compound 
intensifier  is  used,  consisting  of 
three  or  four  cylinders  and  rams, 
nested  in  telescopic  form,  the 
general  arrangement  and  principle 
of  operation  being  the  same  as  in 
the  simple  intensifier  shown  in 
Fig.  6. 

Hydraulic  Accumulator. — A  hy- 
draulic accumulator  is  a  pressure 
regulator  or  governor,  and  bears 
somewhat  the  same  relation  to  a 
hydraulic  system  that  the  flywheel 
does  to  an  engine;  that  is,  it  stores 
up  the  excess  pump  delivery  when 
the  pumps  are  delivering  more  than 
is  being  used,  and  delivers  it  again 
under  pressure  when  the  demand  is 
greater  than  the  supply. 

There  are  two  principal  types  of 
hydraulic  accumulator,  in  one  of 
which  the  ram  is  fixed  or  stationary, 
and  in  the  other  the  cylinder.  The 

latter  type  is  shown  in  Fig.  7.  When  the  delivery  of  the  pumps 
is  greater  or  less  than  required  by  the  machine,  water  enters 
or  leaves  the  cylinder  of  the  accumulator  through  the  pipe  A. 
The  ram  is  thereby  raised  or  lowered,  and  with  it  the  weights 
suspended  by  the  yoke  from  its  upper  end.  The  pressure  in 


Weights 


tr 

A 

^3  Water 


FlG.   7. 


10 


ELEMENTS  OF  HYDRAULICS 


the  system  is  thereby  maintained  constant  and  free  from  the 
pulsations  of  the  pump. 

The  capacity  of  the  accumulator  is  equal  to  the  volume  of  the 
ram  displacement,  and  should  be  equal  to  the  delivery  from  the 
pump  in  five  or  six  revolutions. 

The  diameter  of  the  ram  should  be  large  enough  to  prevent  a 
high  speed  in  descent,  so  as  to  avoid  the  inertia  forces  set  up  by 
sudden  changes  in  speed. 


FIG.  8. 


FIG.  9. 


Hydraulic  Jack. — The  hydraulic  jack  is  a  lifting  apparatus 
operated  by  the  pressure  of  a  liquid  under  the  action  of  a  force 
pump.  Thus  in  Fig.  8  the  hand  lever  operates  the  pump  piston 
B,  which  forces  water  from  the  reservoir  A  in  the  top  of  the  ram 
through  the  valve  at  C  into  the  pressure  chamber  D  under  the 
ram.  The  force  exerted  is  thereby  increased  in  the  direct  ratio 
of  the  areas  of  the  two  pistons.  Thus  if  the  diameter  of  the  pump 
piston  is  1  in.  and  the  diameter  of  the  lifting  piston  or  ram  is  4  in., 


HYDROSTATICS 


11 


the  area  of  the  ram  will  be  sixteen  times  that  of  the  pump  piston. 
If  then  a  load  of,  say,  3  tons  is  applied  to  the  pump  piston  by 
means  of  the  lever,  the  ram  will  exert  an  upward  lifting  force  of 
48  tons. 


Pump  to  Return  Valve  Discharge 
to  Main  System 

FIG.  10. 


Hydraulic  Crane. — The  hydraulic  crane,  shown  in  Fig.  9,  con- 
sists essentially  of  a  ram  and  cylinder,  each  carrying  a  set  of 


12  ELEMENTS  OF  HYDRAULICS 

pulleys.  A  chain  or  rope  is  passed  continuously  over  the  two  sets 
of  pulleys  as  in  the  case  of  an  ordinary  block  and  tackle,  the  free 
end  passing  over  guide  pulleys  to  the  load  to  be  lifted.  When 
water  is  pumped  into  the  cylinder  under  pressure,  the  two  pulley 
blocks  are  forced  apart,  thereby  lifting  the  load  at  the  free  end 
of  the  chain  or  rope. 

Hydraulic  Elevator. — In  hydraulic  installations  two  or  more  of 
these  simple  pressure  machines  are  often  combined,  as  in  the  hy- 
draulic elevator  shown  in  Fig.  10.  In  this  case  the  accumulator 
serves  to  equalize  the  pump  pressure,  making  the  operation  of  the 
system  smooth  and  uniform. 

The  main  valve  for  starting  and  stopping  is  operated,  in  the 
type  shown,  by  the  discharge  pressure,  maintained  by  means  of 
an  elevated  discharge  reservoir.  A  pilot  valve,  operated  from  the 
elevator  cab,  admits  this  low-pressure  discharge  water  to  opposite 
sides  of  the  main  valve  piston  as  desired,  thereby  either  admitting 
high-pressure  water  from  the  pump  and  accumulator,  or  opening 
the  outlet  valve  into  the  discharge.  The  other  details  of  the  in- 
stallation are  indicated  on  the  diagram. 

4.  PRESSURE  ON  SUBMERGED  SURFACES 

Change  of  Pressure  with  Depth. — For  a  liquid  at  rest  in  an 
open  vessel  or  tank,  the  free  upper  surface  is  perfectly  level.  Let 
the  atmospheric  pressure  on  this  surface  be  denoted  by  p.  Then, 
from  Eq.  (3),  the  pressure  p'  at  a  depth  h  below  the  surface 
is  given  by 

Pf  =  P  +  yh. 

Since  the  atmospheric  pressure  is  practically  constant,  the  free 
surface  of  the  liquid  may  be  assumed  as  a  surface  of  zero  pressure 
when  considering  only  the  pressure  due  to  the  weight  of  the  liquid. 
In  this  case  p  =  0,  and  the  pressure  p'  at  any  depth  h,  due  to  the 
weight  of  the  liquid,  becomes 

P'  =  7h.  (7) 

Hence  the  pressure  at  any  point  in  a  liquid  due  to  its  own  weight 
is  directly  proportional  to  the  depth  of  this  point  below  the  free 
upper  surface. 

Moreover,  let  A^l  denote  any  element  of  a  submerged  surface. 
Then  the  pressure  on  it  is 

p'AA 


HYDROSTATICS  13 

Therefore  the  pressure  on  any  element  of  area  of  a  submerged 
surface  is  equal  to  the  weight  of  a  column  of  water  of  cross  section 
equal  to  the  element  considered,  and  of  height  equal  to  the  depth 
of  this  element  below  the  surface. 

Pressure  on  Submerged  Area. — Consider  the  pressure  on  any 
finite  area  A  in  the  side  of  a  tank  a  reservoir  containing  a  liquid 
at  rest  (Fig.  11).  Let  A  A  denote  any  element  of  this  area 
and  x  its  distance  below  the  surface  of  the  liquid.  Then,  by  what 
precedes,  the  pressure  on  this  elementary  area  is 


FIG.  11. 

and  consequently  the  total  pressure  P  on  the  entire  area  A   is 
given  by  the  summation 


But  from  the  ordinary  formula  for  finding  the  center  of  gravity 
of  an  area,  the  distance  XQ  of  the  center  of  gravity  of  A  below  the 
surface  is  given  by 

AxQ  =  Zx&A 
and  consequently 

P  =  7Ax0.  (8) 

Therefore,  the  pressure  of  a  liquid  on  any  submerged  plane  surface 
is  equal  to  the  weight  of  a  column  of  the  liquid  of  cross  section  equal 
to  the  given  area  and  of  height  equal  to  the  depth  of  the  center  of 
gravity  of  this  area  below  the  free  surface  of  the  liquid. 

Center  of  Pressure.  —  The  point  of  application  of  the  resultant 
pressure  on  any  submerged  area  is  called  the  center  of  pressure, 
and  for  any  plane  area  which  is  not  horizontal,  lies  deeper  than  the 


14 


ELEMENTS  OF  HYDRAULICS 


center  of  gravity  of  this  area.  For  instance,  consider  the  water 
pressure  against  a  masonry  dam  with  plane  vertical  face  (Fig.  12). 
By  Eq.  (7)  the  pressure  at  any  point  A  is  proportional  to  the 
depth  of  A  below  the  water  surface.  If,  then,  a  length  AB  is 
laid  off  perpendicular  to  the  wall  and  equal  to  the  depth  of  A 
below  the  surface,  that  is,  AB  =  AO,  then  AB  will  represent  to  a 
certain  scale  the  normal  pressure  at  A.  If  the  same  is  done  at 
various  other  points  of  the  wall,  their  ends,  B,  D,  etc.,  will  lie  in 
a  straight  line  inclined  at  45°  to  the  horizontal.  For  a  portion  of 


FIG. 


the  wall  of  length  b,  the  pressure  acting  on  it  will  then  be  equal  to 
the  weight  of  the  water  prism  OEF,  namely, 

P=7y-2;  (9) 

and  the  center  of  pressure  will  coincide  with  the  center  of  gravity 
of  this  prism.  It  therefore  lies  at  a  distance  of  f h  below  the 
water  surface,  which  is  below  the  center  of  gravity  of  the  rectan- 
gular area  under  pressure  since  the  latter  is  at  a  distance  of  -^ 

from  the  surface. 

General  Formula  for  Center  of  Pressure. — To  obtain  a  general 
expression  for  the  location  of  the  center  of  pressure,  consider  any 
plane  area  inclined  at  an  angle  a  to  the  horizontal  (or  water  sur- 
face) and  subjected  to  a  hydrostatic  pressure  on  one  side.  Let  00' 
denote  the  line  of  intersection  of  the  plane  in  which  the  given  area 
lies  with  the  water  surface  (Fig.  13).  Also,  let  AA  denote  any 
element  of  the  given  area,  h  the  depth  of  this  element  below  the 


HYDROSTATICS  15 

surface,  and  x  its  distance  from  00',  as  indicated  in  the  figure. 
Then  from  Eq.  (7),  the  pressure  AP  acting  on  this  element  is 

AP  =  yhAA, 

and  the  moment  of  this  force  with  respect  to  the  line  00'  is 

xAP  =  yhx  AA. 

Now  let  P  denote  the  total  resultant  pressure  on  the  area  A  and 
xc  the  distance  of  the  point  of  application  of  this  resultant  from 
00',  i.e.,  xc  represents  the  x  coordinate  of  the  center  of  pressure. 
Then  since  the  sum  of  the  moments  of  all  the  elements  of  pressure 


FIG.  13. 

with  respect  to  any  axis  00'  is  equal  to  the  moment  of  their  re- 
sultant with  respect  to  this  axis,  we  have 

2zAP  =  Pxc. 
But  AP  =  yhAA  and  P  =  ZyhAA.     Consequently  this  becomes 

Zyhx&A  =  XcZyhAA. 
Also,  since 

h  =  x  sin  a, 
this  may  be  written 

7  sin  a2x2AA  =  7  sin  axc'2xAA 
or,  cancelling  the  common  factor  7  sin  a, 


The  left  member  of  this  expression  is  by  definition  the  moment  of 
inertia,  7,  of  the  area  A  with  respect  to  the  line  00'  }  that  is 


16  ELEMENTS  OF  HYDRAULICS 

while  by  the  formula  for  the  center  of  gravity  of  any  area  A  we 
also  have 

x0A 


where  x0  denotes  the  x  coordinate  of  the  center  of  gravity  of  A  . 
The  x  coordinate  of  the  center  of  pressure  is  therefore  determined 
by  the  general  formula 

I 
=  Ax0* 

Application.  —  In  applying  this  formula  it  is  convenient  to  use 
the  familiar  relation 

I  =  I,  +  Ad2 
where 

7  =  moment  of  inertia  of  A  with  respect  to  the  axis  00'  ; 
Ig  =  moment  of  inertia  of  A  with  respect  to  a  gravity  axis 

parallel  to  00'; 

d  =  distance  between  these  two  parallel  axes. 
For  example,  in  the  case  of  the  vertical  dam  under  a  hydrostatic 
head  h,  considered  above,  we  have  for  a  rectangle  of  breadth  6  and 
height  h, 

_  bh* 
lg  ~''    12' 

Consequently  the  moment  of  inertia  /  with  respect  to  its  upper 
edge  is 

T         ^' 

: 


and  therefore  the  depth  of  the  center  of  pressure  below  the  sur- 
face is 

bh* 
I  3         2 


which,  of  course,  agrees  with  the  result  obtained  geometrically. 

5.  EQUILIBRIUM  OF  TWO  FLUIDS  IN  CONTACT 

Head  Inversely  Proportional  to  Specific  Weight. — If  two  open 
vessels  containing  the  same  fluid,  say  water,  are  connected  by  a 
tube,  the  fluid  will  stand  at  the  same  level  in  both  vessels  (Fig. 
14).  If  the  two  vessels  contain  different  fluids  which  are  of 


HYDROSTATICS 


17 


different  weights  per  unit  of  volume,  that  is  to  say,  of  different 
specific  gravities,  then  since  the  fluid  in  the  connecting  tube 
must  exert  the  same  pressure  in  either  direction,  the  surface  of 
the  lighter  fluid  will  be  higher  than  that  of  the  heavier. 

For  instance,  let  «i  and  s2  denote  the  specific  gravities  of  the 
two  fluids  in  the  apparatus  shown  in  Fig.  15,  and  let  A  denote 
the  area  of  their  surface  of  contact.  Then  for  equilibrium 


ys2AH 


whence 


Ji  _  82 
H~SI* 


(10) 


The  ratio  of  the  heights  of  the  two  fluids  above  their  surface  of 
separation  is  therefore  inversely  proportional  to  the  ratio  of  their 
specific  gravities. 


FIG.  14. 


FIG.  15. 


Water  Barometer.  —  If  one  of  the  fluids  is  air  and  the  other 
water,  we  have  what  is  called  a  water  barometer.  For  example, 
suppose  that  a  long  tube  closed  at  one  end  is  filled  with  water  and 
the  open  end  corked.  Then  if  it  is  placed  cork  downward  in  a 
vessel  of  water  and  the  cork  removed,  the  water  in  the  tube  will 
fall  until  it  stands  at  a  certain  height  h  above  the  surface  of  the 
water  in  the  open  vessel,  thus  leaving  a  vacuum  in  the  upper  end 
of  the  tube.  The  absolute  pressure  in  the  top  of  the  tube,  A 
(Fig.  16),  is  therefore  zero,  and  at  the  surface,  B,  is  equal  to  the 
pressure  of  the  atmosphere,  or  approximately  14.7  pounds  per 
square  inch.  But  from  Eq.  (3)  we  have 

P     =  P 


where  in  the  present  case 
2 


18 


ELEMENTS  OF  HYDRAULICS 


PB  =  14.7  Ib.  per  square  inch;  PA  =  0;  7  =  62.4  Ib.  per  cubic  foot 
and  by  substitution  of  these  values  we  find  that 

h  =  34  ft.  approximately. 

This  is  the  height,  therefore,  at  which  a  water  column  may  be 
maintained  by  ordinary  atmospheric  pressure.  It  is  therefore 
also  the  theoretical  height  to  which  water  may  be  raised  by  means 
of  an  ordinary  suction  pump.  As  it  is  impossible  in  practice  to 
secure  a  perfect  vacuum,  however,  the  actual  working  lift  for  a 
suction  pump  does  not  exceed  20  or  25  ft. 

Mercury  Barometer. — If  mercury  is  used  instead  of  water, 
since  the  specific  gravity  of  mercury  is  s  =  13.5956,  we  have 

14.7  X  144 
==  13.6  X  62.4  =    °  m*>  aPProximately> 


FIG.  16. 


FIG.  17. 


which  is  accordingly  the  approximate  length  of  an  ordinary 
mercury  barometer. 

Piezometer. — When  a  vessel  contains  liquid  under  pressure, 
this  pressure  is  conveniently  measured  by  a  simple  device  called 
a  piezometer.  In  its  simplest  form  this  consists  merely  of  a  tube 
inserted  in  the  side  of  the  vessel,  of  sufficient  height  to  prevent 
overflow  and  large  enough  in  diameter  to  avoid  capillary  action, 
say  over  1/4  in.  inside  diameter.  The  height  of  the  free  surface 
of  the  liquid  in  the  tube  above  any  point  B  in  the  vessel  then 
measures  the  pressure  at  B  (Fig.  17).  Since  the  top  of  the  tube 
is  open  to  the  atmosphere,  the  absolute  pressure  at  B  is  that 
due  to  a  head  of  h  +  34  ft. 


HYDROSTATICS 


19 


Mercury  Pressure  Gage. — In  general  it  is  convenient  to  use  a 
mercury  column  instead  of  a  water  column,  and  change  the  form 
of  the  apparatus  slightly.  Thus  Fig.  18  shows  a  simple  form  of 
mercury  pressure  gage,  the  difference  in  level,  h,  of  the  two  ends 


FIG.  18. 

of  the  mercury  column  measuring  the  pressure  at  B.  Let  s 
denote  the  specific  weight  of  mercury  and  s'  the  specific  weight 
of  the  fluid  in  the  vessel.  The  pressure  at  any  point  C  in  the 
vessel  is  then 

pc  =  7sh  -  7s'h'.  (u) 


FIG.  19. 

For  example,  if  the  fluid  in  the  vessel  is  water,  then  7  =  62.4, 
s'  =  1,  s  =  13.596,  and  consequently 

pc  =  62.4(13.596/i  -  h'). 


20 


ELEMENTS  OF  HYDRAULICS 


In  case  there  is  a  partial  vacuum  in  the  vessel,  the  gage  may  be 
of  the  form  shown  in  Fig.  19.  The  pressure  on  the  free  surface 
AB  in  the  reservoir  is  then  the  same  as  at  the  top  of  the  barometer 
column,  C,  namely, 

pc  =  14.7  —  ysh. 

6.  EQUILIBRIUM  OF  FLOATING  BODIES 

Buoyancy. — When  a  solid  body  floats  on  the  water  partially 
submerged,  as  in  the  case  of  a  piece  of  timber  or  the  hull  of  a 
ship,  each  element  of  the  wetted  surface  experiences  a  unit  normal 
pressure  of  amount 

p  =  yh 

where  h  denotes  the  depth  of  the  element  in  question  below  the 
water  surface.     Since  the  body  is  at  rest,  the  total  pressure  acting 


=  W-B 


FIG.  20. 

on  the  wetted  surface  together  with  the  weight  of  the  body,  which 
in  this  case  is  the  only  other  external  force,  must  then  form  a  sys- 
tem in  equilibrium.  Since  the  weight  of  the  body  acts  vertically 
downward,  the  water  must  therefore  exert  an  upward  pressure  of 
the  same  amount.  This  resultant  upward  pressure  of  the  water 
is  called  the  buoyant  effort,  or  buoyancy,  and  the  point  of  applica- 
tion of  this  upward  force  is  called  the  center  of  buoyancy.  For 
equilibrium,  therefore,  the  buoyancy  must  be  equal  to  the  weight 
of  the  body  and  act  vertically  upward  along  the  same  line,  since 
otherwise  these  two  forces  would  form  a  couple  tending  to  tip  or 
rotate  the  body  (Fig.  20). 


HYDROSTATICS  21 

Floating  Equilibrium. — To  calculate  the  buoyancy,  suppose 
that  the  solid  body  is  removed  and  the  space  it  occupied  below  the 
water  line  refilled  with  water.  Then  since  the  lateral  pressure  of 
the  water  in  every  direction  must  be  exactly  the  same  as  before, 
the  buoyancy  must  be  equal  to  the  weight  of  this  volume  of  water. 
The  buoyancy  is  therefore  equal  to  the  weight  of  the  volume  of 
water  displaced  by  the  floating  body,  and  the  center  of  buoyancy 
coincides  with  the  center  of  gravity  of  the  displacement.  For 
equilibrium,  therefore,  a  solid  body  must  sink  until  the  weight  of 
the  water  it  displaces  is  equal  to  the  weight  of  the  body,  and  the 
centers  of  gravity  of  the  body  and  its  displacement  must  lie  in 
the  same  vertical. 

These  conditions  also  apply  to  the  case  when  a  body  is  entirely 
submerged.  As  the  density  of  water  increases  with  the  depth,  if 
a  solid  is  slightly  heavier  than  the  water  it  displaces,  it  will  sink 
until  it  reaches  a  depth  at  which  the  density  is  such  that  the 
weight  of  the  water  it  displaces  is  exactly  equal  to  its  own  weight. 

Theorem  of  Archimedes. — If  a  solid  is  heavier  than  the  weight 
of  water  it  displaces,  equilibrium  may  be  maintained  by  suspend- 
ing the  body  in  water  by  a  cord  (Fig.  20),  in  which  case  the  ten- 
sion, T,  in  the  cord  is  equal  to  the  difference  between  the  weight 
of  the  body  and  its  buoyancy,  that  is,  the  weight  of  the  water  it 
displaces.  A  solid  immersed  in  a  liquid  therefore  loses  in  weight 
an  amount  equal  to  the  weight  of  the  liquid  displaced.  This  is 
known  as  the  Theorem  of  Archimedes,  and  was  discovered  by  him 
about  the  year  250  B.  C. 

Physical  Definition  of  Specific  Weight. — Consider  a  solid  com- 
pletely immersed  in  a  liquid,  and  let  V  denote  the  volume  of  the 
solid,  and  7  the  weight  of  a  cubic  unit  of  the  liquid,  say  1  cu.  ft. 
Then  the  buoyancy,  B,  of  the  body  is 

B  =  yV. 

Also,  if  71  denotes  the  weight  of  a  cubic  unit  of  the  solid,  regarded 
as  uniform  and  homogeneous,  its  weight  is 

W  =7iF. 
The  ratio 

£  =  ^  =  s  (») 

B         7 

is  called  the  specific  weight  of  the  solid  with  respect  to  the  liquid 
in  which  it  is  immersed  (compare  Art.  1).  In  general,  the  liquid 
to  which  the  specific  weight  refers  is  assumed  to  be  water  at  a 


22  ELEMENTS  OF  HYDRAULICS. 

temperature  of  39°  F.  The  specific  weight  of  any  substance  is 
then  that  abstract  number  which  expresses  how  many  times 
heavier  it  is  than  an  equal  volume  of  water  at  39°  F.  The  specific 
weight  of  water  is  therefore  unity;  for  lighter  substances  such  as 
wood  or  oil  it  is  less  than  unity;  and  for  heavier  substances  like 
lead  and  mercury  it  is  greater  than  unity. 

Determination  of  Specific  Weight  by  Experiment.  —  The  spe- 
cific weight  of  a  body  may  be  determined  by  first  weighing  it  in 
air  and  again  when  immersed  in  water.  The  actual  weight  of 
the  body  in  air  is  then 

W  =  71  V  =  ysV 

where  s  denotes  its  specific  weight,  and  its  apparent  weight  T 
when  immersed  (Fig.  20)  is 

T  =  W  -  B  =  ysV  -  yV  =  yV(s  -  1), 
that  is, 

T  =  7V(s  -  i).  (13) 

Therefore,  by  division, 

W  s 

T    ~  s  -1 
whence 

W 


S  = 


The  specific  weight  of  a  body  is  therefore  equal  to  its  weight  in 
aif  divided  by  its  loss  in  weight  when  immersed  in  water. 

Application  to  Alloy.  —  If  a  body  is  an  alloy  or  mixture  of  two 
different  substances  whose  specific  weights  are  known,  the  volume 
of  each  substance  may  be  determined  by  weighing  the  body  in 
air  and  in  water.  Thus  let  V\  denote  the  volume,  and  s\  the  spe- 
cific weight,  of  one  substance,  and  F2,  s2  of  the  other.  Then  the 
weight  of  the  body  in  air  is 

W  = 


and  its  apparent  weight  T  when  immersed  is,  from  equation  (13), 
T  =7V1(s1~  1)  +  TF2(s2-  1). 

Solving  these  two  equations  simultaneously  for  V\  and  F2,  the 
result  is 

T  -  (l  -  I)  W 

sj 


-17 


HYDROSTATICS 


23 


en 


Hfl 

This  method  of  determining  relative  volumes  was  invented  by 
Archimedes  in  order  to  solve  a  practical  problem.  Hiero,  King  of 
Syracuse,  had  furnished  a  quantity  of  gold  to  a  goldsmith  to 
be  made  into  a  crown.  When  the  work  was  completed  the  crown 
was  found  to  be  of  full  weight,  but  it  was  suspected  that  the  gold- 
smith had  kept  out  a  considerable  amount  of  gold  and  substi- 
tuted an  equal  weight  of  silver.  To  test  the  truth  of  this  sus- 
picion Archimedes  first  balanced  the  crown  in  air  against  an  equal 
weight  of  gold,  and  then  immersed  both  in  water,  when  the  gold 
was  found  to  outweigh  the  crown, 
proving  the  goldsmith  to  be  dis- 
honest. 

Zero  Buoyancy. — When  a  body 
lies  flat  against  the  bottom  of  a 
vessel  filled  with  water,  fitting  the 
bottom  so  closely  that  no  water  can 
get  under  it,  its  buoyancy  is  zero. 
In  this  case  if  W  denotes  the 
weight  of  the  body,  A  the  area  of 
its  horizontal  cross  section,  and  h 
the  depth  of  water  on  it,  the  force 
T  required  to  lift  it  is  (Fig.  21) 

T  =  W  +  yAh. 

That  is  to  say,  the  force  T  is  the  same  as  would  be  necessary  to 
lift  the  body  itself  and  the  entire  column  of  water  vertically  over 
it. 

This  same  principle  underlies  the  action  of  a  leather  sucker  or 
vacuum  tipped  arrow,  the  fluid  in  that  case  being  air. 


FIG.  21. 


7.  METACENTER 

Stability  of  Floating  Body. — When  a  floating  body  is  shoved 
to  one  side  it  remains  in  this  position  and  is  therefore  in  neutral 
equilibrium  as  regards  lateral  translation.  In  determining  the 
stability  of  a  floating  body  it  is  therefore  only  necessary  to  con- 
sider its  equilibrium  as  regards  rotation. 


24 


ELEMENTS  OF  HYDRAULICS 


After  a  floating  body  has  been  tipped  or  rotated  a  small  amount 
from  its  position  of  equilibrium,  the  buoyancy,  in  general,  no 
longer  passes  through  the  center  of  gravity  of  the  body.  Conse- 
quently the  weight  and  buoyancy  together  form  a  couple  tending 
to  produce  rotation  or  tip  the  body.  If  this  couple  tends  to' 
right  the  body  the  equilibrium  is  stable,  whereas  if  it  tends  to  tip 
it  over  it  is  unstable.  This  evidently  depends  on  the  form  of  the 


FIG.  22. 

wetted  surface,  and  also  on  the  form  of  the  part  immersed  by  the 
rotation. 

Metacenter. — For  example,  consider  a  floating  box  of  rectangu- 
lar cross  section,  immersed  to  a  depth  d  below  the  surface  A  A 
(Fig.  22),  and  suppose  it  is  tipped  by  an  external  couple  until 
the  water  line  becomes  A' A'.  In  this  new  position  the  displace- 
ment is  trapezoidal,  and  the  cen- 
ter of  buoyancy  B  is  the  center  of 
gravity  of  this  trapezoid.  But 
since  the  buoyancy  is  of  the  same 
amount  as  before  the  box  was 
tipped  and  the  triangle  of  immer- 
sion mno  is  equal  to  the  triangle  of 
emersion  opq,  the  lines  AA  and 
A' A'  intersect  on  the  vertical  axis 
CC.  The  intersection  M  of  the  line  of  action  of  the  buoyancy 
with  the  vertical  axis  CC  is  called  the  metacenter.  Evidently 
the  location  of  the  metacenter  depends  on  the  angle  of  tip 
and  is  different  for  each  position.  It  is  also  apparent  that  the 
equilibrium  is  stable  if  the  metacenter  M  lies  above  the  center 
of  gravity  G  of  the  body,  and  unstable  if  M  lies  below  (7.  It  is 
also  shown  in  what  follows  that  the  metacenter  moves  higher  as 


FIG.  23. 


HYDROSTATICS  25 

the  angle  of  tip,  a,  increases.     Its  lowest  position  is  called  the 
true  metacenter. 

Coordinates  of  Metacenter.  —  For  the  special  case  of  the  rec- 
tangular cross  section  shown  in  Fig.  22,  let  x,  y  denote  the  co- 
ordinates of  the  center  of  gravity  of  the  trapezoid,  and  a,  6,  c  the 
lengths  of  three  sides  (Fig.  23).  Then  from  geometry, 

_  c(2a  +  6)  a2  +  ab  +  &2 

=  3  (a  +  6)  '  y~       3  (a  +  6) 

From  Fig.  21  the  sides  a  and  b  of  the  trapezoid  expressed  in  terms 
of  d  and  a  are 

c  c 

a  =  d  —  o  tan  a;   b  =  d  +  ~  tan  a. 

Inserting  these  values  of  a  and  b  in  the  expressions  for  x  and  ?/, 
the  result  is 

c  (3d  -  I  tan  a)  3d2  +  ^  tan2  a 


Also,  from  Fig.  22,  the  total  height  H  of  the  metacenter  above  the 
bottom  of  the  vessel  or  box  is 


/c         \ 
H  =  y  +  (  ^  ~  x  I  cot  a. 


Hence  by  inserting  the  above  values  for  x  and  y  in  this  expression 
for  H  and  reducing,  we  obtain  the  relation 

H  =   2 


The  height  H  therefore  increases  with  a;  that  is,  the  greater  the 
angle  of  tip,  the  higher  the  metacenter  M  .  Moreover,  by  substi- 
tuting a  =  0  in  Eq.  (15)  the  position  of  the  true  metacenter, 
or  limiting  position  of  M,  is  found  to  be  at  a  height  Hf  above  the 
bottom  given  by 


To  prevent  a  ship  from  capsizing,  it  is  necessary  to  so  design  and 
load  it  that  the  height  of  its  center  of  gravity  above  the  bottom 
shall  be  less  than  H'. 

Metacentric  Height.  —  To  consider  the  general  ease  of  equi-- 
librium  of  a  floating  body,  take  a  vertical  cross  section  through 


26  ELEMENTS  OF  HYDRAULICS 

the  center  of  gravity  G  of  the  body  (Fig.  24),  and  suppose  that 
by  the  application  of  an  external  couple  it  is  slightly  tipped  or 
rotated  about  an  axis  OY,  drawn  through  0  perpendicular  to  the 
plane  of  the  paper.  Then  the  volume  displaced  remains  un- 
changed, but  the  center  of  buoyancy  B  is  moved  to  some  other 
point  Br.  To  find  the  metacentric  height  hy,  or  distance  from  the 
center  of  gravity  G  of  the  body  to  the  metacenter  M,  let 
V  =  volume  of  liquid  displaced, 

A  =  cross-sectional  area  of  body  in  plane  of  flotation, 
6  =  distance  from  center  of  gravity  G  to  center  of  buoyancy 

B, 

kv  =  radius  of  gyration   of  area   of  flotation  A   about  the 
axis  OY. 


Then  it  can  be  shown  that1 

hy  =  -y^--b.  (17) 

Similarly,  for  rotation  about  the  axis  OX,  the  metacentric  height 
hx  is  given  by 


where  kx  denotes  the  radius  of  gyration  of  the  area  of  flotation  A 
about  the  axis  OX. 

Evidently  the  metacentric  height  is  greater  for  a  displacement 
about  the  shorter  principal  axis  of  the  section  A.  For  instance, 
it  is  easier  to  make  a  ship  roll  than  to  cause  it  to  tip  endwise  or 
pitch. 

The  locus  of  the  centers  of  buoyancy  for  all  possible  displace- 

1  Webster,  Dynamics  of  Particles,  p.  474.     (Teubner). 


HYDROSTATICS  27 

ments  is  called  the  surface  of  buoyancy,  and  the  two  metacenters 
given  by  Eqs.  (17)  and  (18)  are  the  centers  of  curvature 
of  its  principal  sections. 

Period  of  Oscillation. — When  a  floating  body  is  tipped  and 
then  released,  it  will  oscillate,  or  roll,  with  a  simple  harmonic 
motion.  To  find  the  period  of  the  oscillation,  the  general  ex- 
pression for  the  period  of  oscillation  of  a  solid  body  rotating 
about  a  fixed  axis  may  be  applied,  namely1 


*> 


where    P  =  period  or  time  of  a  complete  oscillation, 
W  =  weight  of  the  body, 
/  =  moment  of  inertia  of  the  body  with  respect  to  the 

axis  of  rotation, 
h  =  distance  from  the  center  of  gravity  of  the  body  to 

the  axis  of  rotation. 

Since  I  =  MK2,  where  M  denotes  the  mass  of  the  body  and  K 
its  radius  of  gyration,  and  also  W  =  Mg,  Eq.  (19)  for  the 
period  may  be  written 

P  =  2,JM^  =  ^.  (20) 

VMgh       Vgh 

Rolling  and  Pitching.  —  In  the  present  case,  consider  rotation 
about  the  two  principal  axes  OX  and  OF  of  the  section  .A  in  the 
plane  of  floatation,  and  let  Kx,  Ky  denote  the  radii  of  gyration 
of  the  solid  with  respect  to  these  axes,  and  Px,  Py  the  correspond- 
ing periods,  or  times  of  performing  a  complete  oscillation  about 
these  axes.  Then  from  Eq.  (20), 

ftrg. 
~  r 


Vgh, 

Substituting  in  these  expressions  the  values  of  hx  and  hy  given  by 
Eqs.  (17)  and  (18),  they  become 

r»  2irKx  .p  271-Kj, 

*»  ~ 


k     ,     ,-     b   \  t././A      b\ 


For  a  body  shaped  like  a  ship,  K  and  k  increase  together,  and 
consequently  the  larger  value  of  k  corresponds  to  the  smaller 
period  P.  A  ship  therefore  pitches  more  rapidly  than  it  rolls. 

1  Slocum,  Theory  and  Practice  of  Mechanics,  p.  302.     (Holt  &  Co.) 


28 


ELEMENTS  OF  HYDRAULICS 


For  further  applications  of  the  metacenter  the  student  is  re- 
ferred to  works  on  naval  architecture. 

APPLICATIONS 

1.  The  ram  of  a  hydraulic  press  is  10  in.  in  diameter  and  the 
plunger  is  2  in.  in  diameter.     If  the  plunger  is  operated  by  a 
handle  having  a  leverage  of  8  to  1,  find  the  pressure  exerted  by 
the  ram,  neglecting  friction,  when  a  force  of  150  Ib.  is  applied  to 
the  handle. 

2.  In  a  hydraulic  press  the  diameter  of  the  ram  is  15  m.  and  of 
the  plunger  is  3/4  in.     The  coefficient  of  friction  may  be  assumed 

as  0.12  and  the  width  of  the  packing 
on  ram  and  plunger  is  0.2  of  their  re- 
spective diameters.  What  pressure 
will  be  exerted  by  the  ram  when  a 
force  of  200  Ib.  is  applied  to  the 
plunger?  , 

3.  Water  in  a  pipe  AB  is  to  be 
kept  at  a  constant  pressure  of  1200 
Ib.  per  square  inch  by  forcing  in  a 
plunger  of  diameter  d  (Fig.  25). 
This  is  operated  by  a  piston  of  di- 
ameter Z),  whose  lower  surface  is  sub- 
jected to  the  pressure  of  a  column  of 

water  75  ft.  high.  Find  the  ratio  of  the  two  diameters  d  and  D. 
4.  In  a  hydraulic  pivot  bearing,  a  vertical  shaft  carrying  a 
total  load  W  is  supported  by  hydraulic  pressure  (Fig.  26).  The 
pivot  is  of  diameter  Z),  and  is  surrounded  by  a  U  leather  packing 
of  width  c.  Show  that  the  frictional  moment,  or  resistance  to 
rotation,  is  given  by  the  relation 


J} 


FIG.  25. 


M  =  2 

where  /*  denotes  the  coefficient  of  friction. 

5.  For  an  ordinary  flat  pivot  bearing  of  the  same  diameter  D 
and  for  the  same  coefficient  of  friction  /z  as  in  the  preceding  prob- 
lem, the  frictional  moment  is  given  by  the  relation1 


M  =  —  »WD. 

o 


Slocum,  Theory  and  Practice  of  Mechanics  (Holt)  p.  194. 


HYDROSTATICS 


29 


Show  that  the  hydraulic  pivot  bearing  is  the  more  efficient  of  the 
two  provided  that 

D 
c<-. 

Calculate  their  relative  efficiency  when  c  =  0.2D. 

6.  An  instrument  for  measuring  the  depth  of  the  sea  consists 
of  a  strong  steel  flask,  divided  into  two  compartments  which  are 
connected  by  a  valve.  The  upper  compartment  is  filled  with  920 
grams  of  distilled  water  and  the  lower  compartment  with  mer- 
cury (Fig.  27).  When  lowered  to  the  bottom,  the  outside  pres- 


_I 


FIG.  26. 


FIG.  27. 


sure  forces  the  sea  water  through  a  small  opening  in  the  side  of 
the  flask  and  thereby  forces  the  mercury  through  the  valve  into 
the  upper  compartment.  Assuming  that  the  depth  of  the  sea  in 
certain  parts  of  the  Pacific  ocean  is  9429  meters,  and  that  the 
ratio  of  the  densities  of  distilled  and  salt  water  is  35:36,  find  how 
many  grams  of  mercury  enter  the  upper  compartment.1  The 
modulus  of  compressibility  of  water  is  0.000047,  that  is,  an  in- 
crease in  pressure  of  one  atmosphere  produces  this  decrease  in 
volume. 

7.  A  hydraulic  jack  has  a  3-in.  ram  and  a  3/4-in.  plunger.  If 
the  leverage  of  the  handle  is  10  to  1,  find  what  force  must  be 
applied  to  the  handle  to  lift  a  weight  of  5  tons,  assuming  the  effi- 
ciency of  the  jack  to  be  75  per  cent. 

1  Wittenbauer,  Aufgaben  aus  der  Technischen  Mechanik,  Bd.  III. 


30  ELEMENTS  OF  HYDRAULICS 

8.  A  hydraulic  intensifier  is  required  to  raise  the  pressure  from 
600  Ib.  per  square  inch  to  2500  Ib.  per  square  inch  with  a  stroke 
of  3  ft.  and  a  capacity  of  4  gal.     Find  the  required  diameters  of 
the  rams. 

9.  In  a  hydraulic  intensifier  like  that  shown  in  Fig.  6,  the 
diameters  are  2  in.,  5  in.  and  8  in.,  respectively.     If  water  is  sup- 
plied to  the    large  cylinder  at  a  pressure  of  500  Ib.  per  square 
inch,  find  the  pressure  at  the  high  pressure  outlet. 

10.  How  would  the  results  of  the  preceding  problem  be  modi- 
fied if  the  frictional  resistance  of  the  glands,  or  packing,  is  taken 
into  account,  assuming  that  the  frictional  resistance  of  one  stuff- 
ing box  is  0.05  pd,  where  p  denotes  the  water  pressure  in  pounds 
per  square  inch,  and  d  is  the  diameter  of  the  ram  in  inches? 

11.  A  hydraulic  crane  has  %  ram  10  in.  in  diameter  and  a 
velocity  ratio  of  1 : 12,  that  is,  the  speed  of  the  lift  is  twelve  times 
the  speed  of  the  ram.    Assuming  the  efficiency  of  the  crane  to  be 
50  per  cent.,  find  what  load  it  will  lift  with  a  water  pressure  of 
1500  Ib.  per  square  inch. 

12.  A  hydraulic  crane    has  a  velocity  ratio  of  1:9  and  is 
required  to  lift  a  load  of  4  tons.     Find  the  required  size  of  the 
ram  for  a  pressure  in  the  mains  of  750  Ib.  per  square  inch,  a  loss 
of  head  due  to  friction  of  75  Ib.  per  square  inch,  and  a  mechanical 
efficiency  of  70  per  cent. 

13.  How  many  foot  pounds  of  work  can  be  stored  up  in  a 
hydraulic  accumulator  having  a  ram  10  in.  in  diameter  and  a 
lift  of  12  ft.,  with  a  water  pressure  of  800  Ib.  per  square  inch? 

14.  Find  the  energy  stored  in  an  accumulator  which  has  a  ram 
10  in.  in  diameter,  loaded  to  a  pressure  of  1000  Ib.  per  square  inch, 
and  having  a  stroke  of  25  ft.     If  the  full  stroke  is  made  in  1 
minute  find  the  horsepower  available  during  this  time. 

15.  The  stroke  of  a  hydraulic  accumulator  is  fifteen  times  the 
diameter  of  the  ram  and  the  water  pressure  is  1200  Ib.  per  square 
inch.     Find  the  diameter  of  the  ram  for  a  capacity  of  125  horse- 
power minutes. 

16.  The  ram  of  a  hydraulic  accumulator  is  20  in.  in  diameter, 
the  stroke  25  ft.,  and  the  water  pressure  1050  pounds  per  square 
inch.     If  the  work  during  one  full  downward  stroke  is  utilized  to 
operate  a  hydraulic  crane  which  has  an  efficiency  of  50  per  cent, 
and  a  lift  of  35  ft.,  find  the  load  raised. 

17.  An  accumulator  is  balanced  by  means  of  a  chain  of  length 
I  passing  over  two  pulleys  A  and  B  (Fig.  28)  and  carrying  a  count- 


HYDROSTATICS 


31 


erweight  W  equal  to  the  total  weight  of  the  chain.  Find  the 
distance  apart  of  the  pulleys  and  the  required  weight  of  chain  per 
unit  of  length  in  order  that  this  arrangement  may  balance  the 
difference  in  pressure  during  motion. 


B 


FIG.  28. 

Hint. — Let  A  denote  the  area  of  the  ram  and  w  the  weight  of 
the  chain  per  unit  of  length.  Then  for  the  dimensions  shown  in 
the  figure,  we  have  the  relations 

wx  —  wz  =  yAc, 
x  +  y  +  z  =  I, 
x  +  h  =  b  +  c, 
yAc 


whence 


w  = 


2c  +  2b  -  2h  - 


y 


18.  A  hydraulic  accumulator  has  a  ram  15  in.  in  diameter  and 
carries  a  load  of  60  tons.     Assuming  the  total  frictional  resistance 
to  be  3  tons,  find  the  required  water  pressure  when  the  load  is 
being  raised  and  when  it  is  being  lowered. 

19.  Show  that  the  depth  of  the  center  of  pressure  below  the  sur- 
face for  a  vertical  rectangle  of  breadth  b  and  depth  d,  with  upper 
edge  immersed  to  a  depth  hi  and  lower  edge  to  a  depth  h2  (Fig.  29) 
is  given  by  the  equation 

2A23  - 
xc=- 


32 


ELEMENTS  OF  HYDRAULICS 


20.  Show  that  the  center  of  pressure  for  a  vertical  plane  tri- 
angle with  base  horizontal  and  vertex  at  a  distance  hi  below  the 
surface  (Fig.  30)  is  given  by  the  equation 


Xc 


u 

T  i 

i      i 

1  i 

FIG.  29. 

21.  From  the  results  of  the  preceding  problem  show  that  if  the 
vertex  of  the  triangle  lies  in  the  surface,  the  depth  of  the  center  of 
pressure  is 

x,  =  -  d, 
and  if  the  base  of  the  triangle  lies  in  the  surface 


FIG.  30. 

22.  Show  that  the  depth  of  the  center  of  pressure  below  the  sur- 
face for  a  vertical  circular  area  of  radius  r,  immersed  so  that  its 
center  lies  at  a  depth  h  below  the  surface  is  given  by 


HYDROSTATICS 


33 


FIG.  31. 


23.  A  circular  opening,  2  ft.  in  diameter  in  the  vertical  side  of 
a  tank  is  closed  by  a  circular  cover  held  on  by  two  bolts,  one  14 
in.  above  the  center  of  the  cover  and  the  other  14  in.  below  its 
center.     When  water  stands  in  the  tank  at  a  level  of  20  ft.  above 
the  center  of  the  opening,  find  the  stress  in  each  bolt. 

24.  A  pipe  of  4  ft.  inside  diameter  flows  just  full,  and  is  closed 
by  a  valve  in  the  form  of  a 

flat  circular  plate  balanced  on 
a  horizontal  axis.  At  what 
distance  from  the  center 
should  the  axis  be'  placed  in 
order  that  the  valve  may  bal- 
ance about  it? 

25.  An  automatic  movable 
flood  dam,  or  flashboard,  is 
made  of  timber  and  pivoted 
to  a  back  stay  at  a  certain 

point  C,  as  shown  in  Fig.  31.  The  point  C  is  so  located  that  the 
dam  is  stable  provided  the  water  does  not  rise  above  a  certain 
point  A,  but  when  it  rises  above  this  point  the  dam  automatic- 
ally tips  over.  Determine  where  the  point  C  should  be  located. 

26.  An  opening  in  a  reservoir  wall  is  closed  by  a  plate  2-1/2  ft. 
square,  hinged  at  the  upper  edge,  and  inclined  at  60°  to  the  hori- 
zontal.    The  plate  weighs  250 
lb.,  and  is  raised  by  a  vertical 
chain  attached  to  the  middle 
point  of  its  lower  edge.     If 
the  center  of  the  plate  is  15 
ft.    below    the    surface,    find 
the  pull  on  the  chain  required 
to  open  it. 

27.  A  rectangular  cast-iron 
sluice  gate  in  the  bottom  of  a 
dam  is  3  ft.  high,  4  ft.  wide 
and  3  in.  thick.  The  head 

of  water  on  the  center  of  the  gate  is  35  ft.  Assuming  the  coeffi- 
cient of  friction  of  the  gate  on  the  slides  to  be  1/4,  and  that  there 
is  no  water  on  the  lower  side  of  the  gate,  find  the  force  required 
to  lift  it.  Weight  of  cast  iron  is  450  lb.  per  cubic  foot. 

28.  Flow  from  a  reservoir  into  a  pipe  is  shut  off  by  a  flap  valve, 
as  shown  in  Fig.  32.     The  pivot  A  is  so  placed  that  the  weight  of 


FIG.  32. 


34  ELEMENTS  OF  HYDRAULICS 

the  valve  and  arm  balance  about  this  point.  Calculate  the  pull  P 
in  the  chain  required  to  open  the  valve  for  the  dimensions  given 
in  the  figure. 

29.  The  waste  gate  of  a  power  canal  is  8  ft.  high  and  5  ft.  wide, 
and  when  closed  there  is  a  head  of  10  ft.  of  water  on  its  center. 
If  the  gate  weighs  1000  Ib.  and  the  coefficient  of  friction  between 
gate  and  seat  is  0.4,  find  the  force  required  to  raise  it. 

30.  A  lock  gate  is  30  ft.  wide  and  the  depth  of  water  on  the 
two  sides  is  28  ft.  and  14  ft.  respectively.     Find  the  total  pressure 
on  the  gate  and  its  point  of  application. 

31.  A  lock  is  20  ft.  wide  and  is  closed  by  two  gates,  each  10  ft. 
wide.     If  the  depth  of  water  on  the  two  sides  is  16  ft.  and  4  ft. 
respectively,  find  the  resultant  pressure  on  each  gate  and  its 
point  of  application. 

32.  A  dry  dock  is  60  ft.  wide  at  water  level  and  52  ft.  wide  at 
floor,  which  is  40  ft.  below  water  level.     The  side  walls  have  a 
straight  batter.     Find  the  total  pressure  on  the  gates  and  its 
point  of  application  when  the  gates  are  closed  and  the  dock  empty. 

33.  A  concrete  dam  is  6  ft.  thick  at  the  bottom,  2  ft.  thick  at 
top  and  20  ft.  high.     The  inside  face  is  vertical  and  the  outside 
face  has  a  straight  batter.     How  high  may  the  water  rise  without 
causing  the  resultant  pressure  on  the  base  to  pass  more  than 
6  in.  outside  the  center  of  the  base? 

Note. — A  dam  may  fail  either  by  overturning  or  by  sliding. 
In  general,  however,  if  a  well-laid  masonry  dam  is  stable  against 
overturning  it  will  not  fail  by  sliding  on  a  horizontal  joint.  To 
prevent  sliding  on  the  base,  an  anchorage  should  be  provided  by 
cutting  steps  or  trenches  in  the  foundation  if  it  is  of  rock,  or  in 
the  case  of  clay  or  similar  material  by  making  the  dam  so  massive 
that  the  angle  which  the  resultant  pressure  on  the  base  makes 
with  the  vertical  is  less  than  the  angle  of  friction. 

In  designing  dams  it  is  customary  to  proportion  the  section  so 
that  the  resultant  pressure  on  any  horizontal  joint  shall  fall  within 
the  middle  third  of  the  joint.  If  this  condition  is  satisfied  there 
will  be  no  danger  of  tensile  stresses  developing  in  the  face  of  the 
dam.1 

If  water  is  allowed  to  seep  under  a  dam,  it  will  exert  a  lifting 
effort  equal  to  the  weight  of  a  column  of  water  of  height  equal  to 
static  head  at  this  point.  To  secure  stability  it  is  therefore 

1  Slocum  and  Hancock,  Strength  of  Materials  (Ginn),  Revised  Edition, 
p.  220. 


HYDROSTATICS 


35 


WASTE  WEIR 

--i-x^r        MAXIMUM  SECTION 
3?3?5i5f£&'HES?  *  °  'i*6"'P" 


Copper  strip1  ^Steel  bor^Inspection-uiell  Vyclopecm 


HORIZONTAL  SECTION  AT  EXPANSION-JOINT 


>lssumj£race:  of&re  \ 


KENSICO  DAM 


FIG.  33.— Catskill  Aqueduct  System. 


36 


ELEMENTS  OF  HYDRAULICS 


essential  to  prevent  seepage  by  means  of  a  cut-off  wall,  as  indi- 
cated in  Figs.  33  and  34. 

In  investigating  the  stability  of  a  dam,  however,  the  best  prac- 
tice provides  for  accidental  seepage  by  making  allowance  for  an 
upward  pressure  on  the  base  due  to  a  hydrostatic  head  of  two- 
thirds  the  actual  depth  of  water  back  of  the  dam. 

34.  Figure  33  shows  a  typical  section  of  the  Kensico  Dam,  form- 
ing part  of  the  Catskill  Water  System  of  the  City  of  New  York. 
The  Kensico  Reservoir  covers  2218  acres,  with  a  shore  line  40 


£120 ft.  above  flow 


'  berm 


Drainage  u/eff- 


OLIVE  BRIDGE  DAM 

MAXIMUM  MASONRY  SECTION 

FIG.  34. — Catskill  Aqueduct  System. 

x 

miles  in  length,  and  has  a  storage  capacity  of  38,000,000,000 
gal.  The  dimensions  of  the  main  dam  are  length  1843  ft.; 
height  300  ft. ;  thickness  at  base  230  ft. ;  thickness  at  top  28  ft. 

Investigate  the  stability  of  this  dam  in  accordance  with  the 
conditions  stated  in  the  note  to  Problem  33. 

35.  Figure  34  shows  a  section  of  the  Olive  Bridge  Dam  and  typi- 
cal dyke  section  of  the  Ashokan  Reservoir,  which  forms  part  of  the 
Catskill  Water  System  of  the  City  of  New  York.  This  reservoir 
covers' 8 180  acres,  with  a  shore  line  40  miles  in  length  and  a  stor- 
•age  capacity  of  132,000,000,000  gal.  The  principal  dimensions 


HYDROSTATICS 


37 


of  the  main  dam  are,  length  4650  ft. ;  height  220  ft. ;  thickness  at 
base  190  ft. ;  thickness  at  top  23  ft. 

Investigate  the  stability  of  this  dam  as  in  the  preceding 
problem. 

36.  In  the  $25,000,000  hydraulic  power  development  on  the 
Mississippi  river  at  Keokuk,  Iowa,  the  dam  proper  is  4650  ft. 
long,  with  a  spillway  length  of  4278  ft.  The  power  plant  is 
designed  for  an  ultimate  development  of  300,000  h.p.,  and  consists 
of  vertical  shaft  turbines  and  generators  in  units  of  10,000  h.p. 
each.  Transmission  lines  convey  the  current  at  110,000  volts  to 
St.  Louis,  137  miles  distant,  and  to  other  points.1 


Operating 
Mechanism''' 


H.W.EI.S2S 


.eGage  Lamp 


FIG.  35. — Section  of  side  walls  of  lock  at  Keokuk,  Iowa,  showing  valves 
for  intake  and  discharge  conduits. 

A  notable  feature  of  the  plant  is  the  ship  lock  which  is  of  unus- 
ual size  for  river  navigation,  the  lock  chamber  being  400  ft.  long 
by  110  ft.  wide  with  a  single  lift  of  from  30  to  40  ft.,  the  total 
water  content  of  the  lock  when  full  being  about  2,200,000  cu.  ft. 
The  locks  at  Panama  are  the  same  width  but  the  maximum  lift 
on  the  Isthmus  is  32  ft.,  the  average  lift  being  about  28  ft.  Find 
the  maximum  pressure  on  the  lock  gates  at  Keokuk  and  its  point 
of  application.  (See  frontispiece.) 

37.  The  side  walls  of  the  Keokuk  lock  are  monolithic  masses 
of  concrete,  with  a  base  width  of  33  ft.,  a  top  width  of  8  ft.,  and 

1  Eng.  News,  Sept.  28,  1911. 


38  ELEMENTS  OF  HYDRAULICS 

an  outside  batter  of  1 :1.5,  as  shown  in  Fig.  35.  If  the  water 
stands  48  ft.  above  the  floor  of  the  lock  on  the  inside  and  8  ft. 
on  the  outside,  find  the  point  where  the  resultant  pressure  on  the 
side  walls  intersects  the  base,  neglecting  the  weight  of  the  road- 
way on  top  and  the  arches  which  support  it. 

38.  The  lower  lock  gates  at  Keokuk  are  of  the  mitering  type,  as 
shown  in  Figs.  36  and  37,  and  are  very  similar  to  those  in 
the  Panama  canal  locks.  The  gates  are  49  ft.  high  and  each  leaf 
consists  of  13  horizontal  ribs  curved  to  a  radius  of  66  ft.  4-3/4  in. 
on  the  center  line,  framed  together  at  the  ends  by  the  quoin  and 


FIG.  36. — Steel  miter  gates  for  lower  end  of  lock,  showing  buoyancy 

chamber. 

miter  posts,  and  also  having  seven  lines  of  intermediate  framing. 
The  chord  length  over  the  posts  is  66  ft.  4-3/4  in.  and  the  rise  of 
the  curve  is  10  ft.  8-1/2  in.1 

Each  leaf  contains  a  buoyancy  chamber  to  relieve  the  weight 
on  the  top  hinge.  This  consists  of  a  tank  of  about  3840  cu.  ft. 
capacity,  placed  between  the  curve  of  the  face  and  the  chord  line 
of  the  bracing.  The  total  weight  of  each  gate  in  air  is  about  240 
tons.  Find  how  much  the  buoyancy  chamber  relieves  the  weight 
on  the  top  hinge. 

39.  The  upper  gates  of  the  Keokuk  lock  are  of  a  floating  type 
never  before  used,  and  consist  essentially  of  floating  tanks  moving 
in  vertical  guides  and  sinking  below  the  level  of  the  sill  (Fig.  38). 

1  Eng.  News,  Nov.  13,  1913. 


HYDROSTATICS 


39 


To  close  the  lock,  compressed  air  is  admitted  to  an  open-bottom 
chamber  in  the  gate,  which  forces  out  the  water  and  causes  the 
gate  to  rise.  To  open  the  lock,  the  air  in  this  chamber  is  al- 


Section  of  Gate  erf 
Center  (E-F) 

e"'-Hsw*  Sectional    Plan  C'D 

FIG.  37. — Detail  of  steel  miter  gates  for  lower  end  of  lock. 

lowed  to  escape,  when  the  weight  of  the  gate  sinks  it  to  its  lower 
position. 

The  flotation  of  the  gate  is  controlled  by  two  closed  displace- 


40 


IT   tj 


ELEMENTS  OF  HYDRAULICS 
\ 

^ 


HYDROSTATICS  41 

ment  chambers,  one  at  each  end,  and  one  open  buoyancy  cham- 
ber. Each  of  the  former  is  42  ft.  long,  4  ft.  deep  and  16  ft.  wide. 
The  buoyancy  chamber  is  2-1/2  ft.  high  beneath  the  displacement 
chambers  and  6-1/2  ft.  high  in  the  28-ft.  space  between  them,  its 
capacity  being  6000  cu.  ft. 

With  the  gate  floating  and  its  bottom  just  clear  of  the  sill,  the 
weight  of  the  part  above  water  is  190  tons,  which  is  increased  by 
the  ballast  in  the  displacement  chambers  to  210  tons.  The  dis- 
placement of  the  submerged  part  of  the  gate  is  12  tons  so  that  the 
buoyant  effort  required  is  198  tons. 

Find  the  equivalent  displacement  in  cubic  feet,  from  this  result 
subtract  the  volume  of  the  displacement  chambers,  and  then  find 
the  required  air  pressure  in  the  buoyancy  chamber. 

In  raising  the  gate  it  is  actually  found  that  this  pressure  varies 
from  2  Ib.  per  square  inch  to  as  high  as  12  Ib.  per  square  inch 
when  the  gate  is  leaving  its  lower  seat. 

40.  A  gas  tank  is  fitted  with  a  mercury  gage  as  shown  in  Fig. 
18.     The  height  h  of  the  mercury  column  is  20  in.     Find  the 
excess  of  pressure  in  the  tank  above  atmospheric. 

41.  A  piece  of  lead  weighs  20  Ib.  in  air.     What  will  be  its 
apparent  weight  when  suspended  in  water,  assuming  the  specific 
weight  of  lead  to  be  11.4? 

42.  A  pail  of  water  is  placed  on  a  platform  scales  and  found  to 
weigh  12  Ib.     A  6-lb.  iron  weight  is  then  suspended  by  a  light 
cord  from  a  spring  balance  and  lowered  into  the  water  in  the  pail 
until  completely  immersed.     Find  the  reading  on  the  spring 
balance  and  on  the  platform  scales., 

43.  A  brass  casting  (alloy  of  copper  and  zinc)  weighs  200  Ib. 
in  air  and  175  Ib.  in  water.     If  the  specific  weight  of  copper  is 
8.8  and  of  zinc  is  7,  how  many  pounds  of  each  metal  does  the 
casting  contain? 

44.  One  end  of  a  wooden  pole  12  ft.  long,  floats  on  the  water 
and  the  other  end  rests  on  a  wall  so  that  2  ft.  project   inward 
beyond  the  point  of  support  (Fig.  39).     If  the  point  of  support 
is  18  in.  above  the  water  surface,  find  how  much  of  the  pole  is 
immersed. 

45.  A  floating  platform  is  constructed  of  two  square  wooden 
beams  each  16  ft.  long,  one  18  in.  square  and  the  other  1  ft. 
square.     On  these  is  laid  a  platform  of  2-in.  plank,  10  ft.  wide. 
Find~  where  a  man  weighing  160  Ib.  must  stand  on  the  platform 
to  make  it  float  level,  and  how  high  its  surface  will  then  be  above 


42 


ELEMENTS  OF  HYDRAULICS 


the  water  (Fig.  40).     The  weight  of  timber  may  be  assumed 
as  50  Ib.  per  cubic  foot. 

46.  A  piece  of  timber  4  ft.  long  and  4  in.  square  has  a  weight 
W  attached  to  its  lower  end  so  that  it  floats  in  water  at  an  angle 
of  45°  (Fig.  41).     Find  W. 

47.  A  rectangular  wooden  barge  is  30  ft.  long,  12  ft.  wide  and 
4  ft.  deep,  outside  measurement,  and  is  sheathed  with  plank 


FIG.  39. 

3  in.  thick,  the  frame  weighing  half  as  much  as  the  planking. 
Find  the  position  of  the  water  line  when  the  barge  floats  empty, 
and  also  the  load  in  tons  it  carries  when  the  water  line  is  1  ft.  from 
the  top.  Assume  the  weight  of  wood  as  50  Ib.  per  cubic  foot. 

48.  A  prismatic  wooden  beam  10  ft.  long,  1  ft.  wide  and  6  in. 
thick  floats  flat  on  the  water  with  4  in.  submerged  and  2  in.  above 
water.  Find  its  specific  weight. 


FIG.  40. 

49.  A  dipper  dredge  weighs  1200  tons  and  floats  on  an  even 
keel  with  bucket  extended  and  empty.  When  the  bucket 
carries  a  load  of  3  tons  at  a  distance  of  50  ft.  from  the  center  line 
of  the  scow,  a  plumb  line  15  ft.  long,  suspended  from  a  vertical 
mast,  swings  out  5  in.  Find  the  metacentric  height. 


HYDROSTATICS 


43 


50.  A  steamer  is  of  14,000  tons  displacement.  When  its  life 
boats  on  one  side  are  filled  with  water,  a  plumb  line  20  ft.  long 
suspended  from  a  mast  is  found  to  swing  out  9-1/2  in.  If  the 


FIG.  41. 


total  weight  of  water  in  the  boats  is  75  tons  and  their  distance 
from  the  center  line  of  the  vessel  is  25  ft.,  find  the  period  with 
which  the  ship  will  roll. 


SECTION  II 


HYDROKINETICS 
8.  FLOW  OF  WATER  FROM  RESERVOIRS  AND  TANKS 

Stream  Line. — In  the  case  of  a  flowing  liquid,  the  path  fol- 
lowed by  any  particle  of  the  liquid  in  its  course  is  called  a  stream 
line.  In  particular,  if  a  reservoir  or  tank  is  filled  with  water  and 
a  small  opening  is  made  in  one  side  at  a  depth  h  below  the  surface, 
the  water  flows  out  with  a  certain  velocity  depending  on  the 

depth,  or  head,  h.  Since  the  par- 
ticles of  water  flowing  out  con- 
verge at  the  opening,  the  stream 
lines  inside  the  vessel  are,  in 
general,  comparatively  far  apart, 
but  become  crowded  more  closely 
together  at  the  orifice. 

Liquid  Vein. — Under  the  condi- 
tions just  considered,  suppose  that 
a  closed  curve  is  drawn  in  any 
horizontal  cross  section  of  the 
vessel  and  through  each  point  of 
the  closed  curve  draw  a  stream 
line.  The  totality  of  all  these 
stream  lines  will  then  form  a  tube, 
called  a  liquid  vein  (Fig.  42). 
From  the  definition  of  a  stream 

line  it  is  evident  that  the  flow  through  such  a  tube  or  vein  is  the 
same  as  though  it  were  an  actual  material  tube.  In  particu- 
lar, the  same  amount  of  liquid  will  flow  through  each  cross 
section  of  the  vein  and  therefore  the  velocity  of  flow  will  be 
greatest  where  the  cross  section  of  the  vein  is  least,  and  vice  versa. 
Ideal  Velocity  Head. — In  any  particular  vein  let  v  denote  the 
velocity  of  flow  at  a  distance  h  below  the  surface,  and  Q  the 
quantity  of  water  per  second  flowing  through  a  cross  section  of 
the  vein  at  this  depth.  Then  the  weight  of  water  flowing  through 
the  cross  section  per  second  is  yQ  and  its  potential  energy  at  the 

44 


1: 


FIG.  42. 


HYDROKINETICS  45 

height  h  is  yQh.     The  kinetic  energy  of  this  quantity  of  water 

jQv2 
flowing   at  the  velocity  v  is  —~ — •     Therefore  by  equating  the 

potential  energy  lost  to  the  kinetic  energy  gained  and  neglecting 
all  frictional  and  other  losses  we  have 


whence 

V  =     V2gE  (22) 

This  relation  may  also  be  written  in  the  form 

k=     2g 

The  quantity  h  is  therefore  called  the  ideal  velocity  head,  since 
it  is  the  theoretical  head  required  to  produce  a  velocity  of  flow  v. 
Torricelli's  Theorem. — The  relation 


v  =  V2gh 

is  known  as  Torricelli's  Theorem.  Expressed  in  words,  it  says 
that  the  ideal  velocity  of  flow  under  a  static  head  h  is  the  same 
as  would  be  acquired  by  a  solid  body  falling  in  a  vacuum 
from  a  height  equal  to  the  depth  of  the  opening  below  the  free 
surface  of  the  liquid. 

Actual  Velocity  of  Flow. — The  viscosity  of  the  liquid,  as  well  as 
the  form  and  dimensions  of  the  opening,  have  an  important  effect 
in  modifying  the  discharge. 

Considering  viscosity  first,  its  effect  is  to  reduce  the  velocity 
of  the  issuing  liquid  below  the  ideal  velocity  given  by  the  relation 
v  =  ^2gh.  It  is  therefore  necessary  to  modify  this  relation  so  as 
to  conform  to  experiment  by  introducing  an  empirical  constant 
called  a  velocity  coefficient.  Denoting  this  coefficient  by  Cv,  the 
expression  for  the  velocity  becomes 

v  =  Cv  V2gh.  (23) 

For  water  the  value  of  the  velocity  (or  viscosity)  coefficient  for  an 
orifice  or  a  nozzle  is  approximately  Cv  =  0.97. 

Contraction  Coefficient. — In  the  case  of  flow  through  an  orifice 
or  over  a  weir,  the  oblique  pressure  of  the  water  approaching 
from  various  directions  causes  a  contraction  of  the  jet  or  stream 
so  that  the  cross  section  of  the  jet  just  outside  the  orifice  is  some- 


46 


ELEMENTS  OF  HYDRAULICS 


what  less  than  the  area  of  the  opening.  Consequently  the  dis- 
charge is  also  less  than  it  would  be  if  the  jet  were  the  full  size  of  the 
opening. 

If  the  area  of  the  orifice  is  denoted  by  A,  the  area  of  the  jet  at 
the  contracted  section  will  be  some  fraction  of  this  amount,  say 
CCA ,  where  Cc  is  an  empirical  constant  called  a  contraction  coeffi- 
cient, which  must  be  determined  experimentally  for  openings  of 
various  forms  and  dimensions. 

Efflux  Coefficient. — Taking  into  account  both  the  viscosity  of 
the  liquid  and  the  contraction  of  the  jet,  the  formula  for  dis- 
charge becomes 

Q  =  actual  velocity  X  area  of  jet 
=  (Cv  V2<//0_X  (CcA) 
=  CVCCA  -\l2gh, 

where  A  denotes  the  area  of  the  orifice.  Since  there  is  no  object 
in  determining  Cv  and  Cc  separately,  they  are  usually  replaced  by 


~ 


_  _^_  __  _L^b^_ 


FIG.  43. 


FIG.  44. 


a  single  empirical  constant  K  -  CVCC,  called  the  coefficient  of 
efflux,  or  discharge.  In  general,  therefore,  the  formula  for  the 
actual  discharge  becomes 


Q  =  KA  V2gh. 


(24) 


Effective  Head.— The  head  h  may  be  the  actual  head  of  water 
on  the  orifice;  or  if  the  vessel  is  closed  and  the  pressure  is  produced 
by  steam  or  compressed  air,  the  effective  head  is  the  height  to 
which  the  given  pressure  would  sustain  a  column  of  water. 

The  height  of  the  equivalent  water  column  corresponding  to 
any  given  pressure  may  be  determined  by  calculating  the  weight 


HYDROKINETICS 


47 


of  a  column  of  water  1  ft.  high  and  1  sq.  in.  in  section,  from  which 
it  is  found  that 

i  ft.  head  =  0.434  Ib.  /in.2  pressure, 
and  conversely, 

i  lb./in.2  pressure  =  2.304  ft.  head. 

For  an  orifice  in  the  bottom  of  a  vessel,  the  head  h  is  of  course 
the  same  at  every  point  of  the  opening,  but  if  the  orifice  is  in  the 
side  of  the  vessel,  the  head  h  varies  with  the  depth.  However,  if 
the  depth  of  the  opening  is  small  in  comparison  with  h,  as  is  fre- 
quently the  case,  the  head  may  be  assumed  to  be  constant  over 
the  entire  orifice  and  equal  to  the  distance  of  its  center  of  gravity 
from  the  free  surface  of  the  liquid  (Fig.  43). 

If  an  orifice  is  entirely  submerged,  as  shown  in  Fig.  44,  the 
effective  head  on  it  is  the  difference  in  level  between  the  water 
surfaces  on  the  two  sides  of  the  opening. 

Discharge  from  Large  Rectangular  Orifice. — In  the  case  of  a 
comparatively  large  orifice,  the  effective  head  is  not  the  depth  of 


FIG.  45. 

its  center  of  gravity  below  the  surface,  and  the  discharge  must  be 
determined  in  a  different  manner. 

To  illustrate  the  method  of  procedure  consider  the  particular 
case  of  an  orifice  in  the  form  of  a  rectangle  of  breadth  6,  the  upper 
and  lower  edges  being  horizontal  and  at  depths  of  h  and  H  re- 
spectively below  the  surface,  as  shown  in  Fig.  45.  Let  this 
rectangle  be  divided  up  into  narrow  horizontal  strips,  each  of 
breadth  b  and  depth  dy.  Then  the  ideal  velocity  of  flow  in  any 
one  of  these  strips  at  a  distance  y  below  the  surface  is  v  = 


48 


ELEMENTS  OF  HYDRAULICS 


and  since  its  area  is  bdy,  the  ideal  discharge  dQ  through  this  ele- 
mentary area  per  second  is 

dQ  =  bdy  V2J/y. 

The  total  discharge  per  second,  Q,  through  the  entire  orifice  is 
therefore 

r* 

-  h%).  (25) 


Q  =  Kb  |    V2gy  dy  =  -Kb 
3 

This  expression  may  also  be  written  in  the  form 

tt.'-.l 


which  makes  it  easier  to  remember  from  analogy  with  the  weir 
formula  which  follows. 

Discharge  of  a  Rectangular  Notch  Weir.  —  If  the  upper  edge  of 
the  rectangular  orifice  just  considered  coincides  with  the  water 
surface,  the  opening  is  called  a  rectangular  notch  weir.  In  this 
case  h  =  0  and  the  preceding  formula  for  discharge  becomes 


Q  =  ^KbH  V^H   =  - KA  Vz^ii 
3  3 


(26) 


where  H  denotes  the  head  on  the 
crest  of  the  weir,  and  A  is  the  area 
of  that  part  of  the  opening  which 
lies  below  the  surface. 


~^-=.      9.  DISCHARGE   THROUGH   SHARP- 
EDGED  ORIFICE 


FIG.  46. 


Contraction  of  Jet. — In  consider- 
ing the  flow  of  water  through  an 
orifice  it  is  assumed  in  what  follows 
that  a  sharp-edged  orifice  is  meant, 
that  is,  one  in  which  the  jet  is  in 
contact  with  the  wall  of  the  vessel 
along  a  line  only  (Fig.  46).  When  this  is  not  the  case,  the  open- 
ing is  called  an  adjutage  or  mouthpiece,  and  the  flow  is  modified, 
owing  to  various  causes,  as  explained  in  Art.  13. 

The  value  of  the  constant  K  in  Eq.  (26)  depends  on  the  form 
of  the  opening  and  also  on  the  nature  of  the  contraction  of  the 
jet.  The  contraction  is  said  to  be  complete  when  it  takes  place 


HYDROKINETICS  49 

on  all  sides  of  the  jet;  that  is  to  say,  when  the  size  of  the  opening 
is  small  in  comparison  with  its  distance  from  the  sides  and  bottom 
of  the  vessel  and  from  the  water  surface.  The  contraction  is 
called  incomplete  when  one  of  more  of  the  edges  of  the  orifice  is 
continuous  with  the  sides  of  the  vessel. 

Complete  Contraction. — For  a  sharp-edged  orifice  with  com- 
plete contraction  the  mean  value  of  the  efflux  coefficient  K  is 

K  =  0.62. 

The  actual  value  of  this  coefficient  varies  slightly  with  the  size 
of  the  orifice  and  effective  head  on  it.  The  value  given,  how- 
ever, is  sufficiently  accurate  for  all  ordinary  practical  calcula- 
tions. More  exact  values  are  given  in  Tables  9  and  10. 

Partial  Contraction. — In  the  case  of  incomplete  contraction, 
let  P  denote  the  entire  perimeter  of  the  orifice,  and  nP  that  frac- 
tion of  the  perimeter  which  experiences  no  contraction.  Then 
denoting  the  coefficient  of  efflux  by  KI,  its  value  as  determined  by 
experiment  for  sharp-edged  orifices  is  as  follows: 

Rectangular  orifice,  KI  =  K(i  +  0.1511)  1  .    . 

Circular  orifice,        KI  =  K(i  +  0.1311)  j 

Assuming  K  =  0.62,  the  following  table  gives  the  corresponding 
values  of  KI  as  determined  from  these  relations.1 


•n  =  1 

n  =  \ 

n  =  I 

Rectangular  orifice 

K!  =0.643 

K!  =0.667 

Kj.  =0.690 

Circular  orifice  

Ki  =0.640 

Ki  =0.660 

^  =0.680 

Velocity  of  Approach. — So  far  it  has  been  assumed  that  the 
effective  head  h  in  the  formula  for  discharge  through  an  orifice, 
namely 

Q  =  KA  V2^, 

is  simply  the  static  head,  measured  from  the  center  of  the  orifice, 
if  it  is  small,  to  the  surface  level.  If  the  velocity  of  approach  is 
considerable,  however,  the  velocity  head  must  also  be  included 
in  the  effective  head.  Thus  let 

A  =  area  of  orifice, 
A'  =  cross  section  of  channel  of  approach, 

V  =  ideal  velocity  corresponding  to  the  total  head  H, 
1  Lauenstein,  Mechanik,  p,  173. 

4 


50  ELEMENTS  OF  HYDRAULICS 

v  =  velocity  of  approach, 
h'  =  velocity  head  =  o-» 

h  =  static  head, 

72 
H  =  effective  head  =  y- 

Since  the  total  flow  through  the  channel  of  approach  must  equal 
the  discharge  through  the  orifice,  we  have 

A'v  =  Q  =  KAV 
whence 

KAV 

-jr' 

Also  the  effective  head  H  —  h  +  h',  or,  expressed  in  terms  of  the 
velocities, 


Substituting  v  =      ,,     in  this  relation,  it  becomes 

IP  _          K*A2V2 
2g  ~  h  ' 
whence 


v  =     /          2^ 


The  expression  for  the  discharge  Q  is  then 


2gh 

(28) 


I  —  K' 

From  this  relation  it  is  evident  that  if  the  area  A  of  the  orifice  is 
small  in  comparison  with  the  cross  section  A'  of  the  channel, 
say  A'  not  less  than  fifteen  times  A,  the  error  due  to  neglecting  the 

(A    \   2 
-pj 

in  Eq.  (28)  may  be  neglected,  in  which  case  the  formula 
for  the  discharge  simplifies  into  the  original  expression  given  by 
Eq.  (26),  namely, 

Q  =  KAV2gh. 


HYDROKINETICS 


51 


10.  RECTANGULAR  NOTCH  WEIRS 

Contracted  Weir. — The  most  common  type  of  weir  consists 
of  a  rectangular  notch  cut  in  the  upper  edge  of  a  vertical  wall, 
and  is  called  a  contracted  weir  (Fig.  47).  In  order  that  the  con- 
traction shall  be  complete,  there  should  be  a  clearance  of  not  less 
than  3h  from  the  sides  of  the  notch  to  the  sides  of  the  channel, 
and  from  the  bottom  of  the  notch  (called  the  crest  of  the  weir) 
to  the  bottom  of  the  channel. 

Suppressed  Weir. — If  the  sides  of  the  notch  are  continuous 
with  the  sides  of  the  channel,  it  is  called  a  suppressed  weir  (Fig. 
48). 


FIG.  47. 


FIG.  48. 


For  both  types  of  weir  it  has  been  found  by  experiment  that 
the  velocity  of  approach  may  be  neglected  when  the  product  bh 
is  less  than  one-sixth  the  cross  section  of  the  channel.  For  a 
suppressed  weir  this  is  equivalent  to  saying  that  the  height  of 
the  weir  crest  above  the  bottom  should  be  at  least  five  times  the 
head  on  the  weir. 

Empirical  Weir  Formulas. — Numerous  experiments  have  been 
made  on  the  flow  of  water  over  weirs  for  the  purpose  of  deriving 
an  empirical  formula  for  the  discharge.  The  most  important  of 
these  results,  including  the  formulas  in  common  use,  are  tabulated 
on  page  52.  Although  these  formulas  apparently  differ  somewhat 
in  form,  they  are  found  in  practice  to  give  results  which  agree  very 
closely. 

Rational  Weir  Formula. — A  rational  formula  for  the  discharge 
over  a  weir  was  derived  in  Art.  8,  as  expressed  by  Eq.  (26), 
namely, 

Q  =  %KAV2gh. 


52 


ELEMENTS  OF  HYDRAULICS 


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4  Hamilton  Smith; 

HYDROKINETICS  53 

For  a  sharp-edged  opening  the  mean  value  of  the  efflux  coefficient 
is  K  =  0.62,  as  stated  in  Art.  9.  In  the  present  case,  therefore, 
KA  =  0.626A,  and  if  b  and  h  are  expressed  in  feet,  the  above 
formula  becomes 

Q  =  §{o.62bh)\/2gih, 

=  3.3bh3^  cu.  ft.  per  sec.  (29) 

It  is  often  convenient  to  express  b  and  h  in  inches,  and  the  dis- 
charge Q  in  cubic  feet  per  minute.  Expressed  in  these  units,  the 
formula  becomes 


or,  reducing  and  simplifying, 

Q  =  o.4bh^  cu.  ft.  per  min.  (30) 

where  b  and  h  are  both  expressed  in  inches.  These  formulas  are 
the  basis  of  many  of  the  weir  tables  used  in  practical  work,  such 
as  Tables  17  and  18  in  this  book. 

11.  STANDARD  WEIR  MEASUREMENTS 

Construction  of  Weir.  —  From  the  experiments  summarized  in 
the  preceding  article  it  was  found  that  any  empirical  weir  formula 
could  only  be  relied  upon  to  give  accurate  results  when  the  con- 
ditions under  which  the  measurement  was  made  were  approxi- 
mately the  same  as  those  under  which  the  formula  was  deduced. 
To  obtain  accurate  results  from  weir  measurements  it  is  therefore 
customary  to  construct  the  weir  according  to  certain  standard 
specifications,  as  follows: 

1.  A   rectangular   notch   weir   is   constructed   with   its  edges~ 
sharply  beveled  toward  the  intake,  as  shown  in  Fig.  49.     The 
bottom  of  the  notch,  called  the  crest  of  the  weir,  must  be  per- 
fectly level  and  the  sides  vertical. 

2.  The  length,  or  width,  of  the  weir  should  be  between  four 
and  eight  times  the  depth  of  water  flowing  over  the  crest  of  the 
weir. 

3.  The  channel  or  pond  back  of  the  weir  should  be  at  least 
50  per  cent,  wider  than  the  notch,  and  of  sufficient  depth  so  that 
the  velocity  of  approach  shall  not  be  over  1  ft.  per  second.     In 
general  it  is  sufficient  if  the  area  bh  is  not  over  one-sixth  the  area 


54 


ELEMENTS  OF  HYDRAULICS 


of  the  channel  section  where  6  denotes  the  width  of  the  notch 
and  h  ,the  head  of  water  on  the  crest. 

4.  To  make  -the  end  contractions  complete  there  must  be  a 
clearance  of  from  2h  to  3h  between  each  side  of  the  notch  and 
the  corresponding  side  of  the  channel. 

5.  The  head  h  must  be  accurately  measured.     This  is  usually 
accomplished  by  means  of  an  instrument  called  a  hook  gage 
(Fig.  50),  located  as  explained  below.     For  rough  work,  however, 


FIG.  49. 

the  head  may  be  measured  by  a  graduated  rod  or  scale,  set  back 
of  the  weir  at  a  distance  not  less  than  the  length  of  the  notch, 
with  its  zero  on  a  level  with  the  crest  of  the  weir  (Fig.  49). 

Hook  Gage. — As  usually  constructed,  the  hook  gage  consists 
of  a  wooden  or  metal  frame  carrying  in  a  groove  a  metallic  sliding 
scale  graduated  to  feet  and  hundredths,  which  is  raised  and  low- 
ered by  means  of  a  milled  head  nut  at  the  top  (Fig.  50).  By  means 
of  a  vernier  attached  to  the  frame,  the  scale  may  be  read  to  thou- 
sandths of  a  foot.  The  lower  end  of  the  frame  carries  a  sharp- 
pointed  brass  hook,  from  which  the  instrument  gets  its  name. 

In  use,  the  hook  gage  is  set  up  in  the  channel  abave  the  weir 
and  leveled  by  means  of  a  leveling  instrument  so  that  the  scale 


HYDROKINETICS 


55 


reads  zero  when  the  point  of  the  hook  is  at  the  exact  level  of  the 
crest  of  the  weir.  The  hook  is  then  raised  until  its  point  just 
reaches  the  surface,  causing  a  distortion  in  the  reflection  of  light 
from  the  surface  of  the  water.  If  slightly  lowered  the  distortion 
disappears,  thus  indicating  the  surface  level  with  precision. 
The  reading  of  the  vernier  on  the  scale 
then  gives  the  head  on  the  crest  to 
thousandths  of  a  foot. 

Location  of  Hook  Gage. — To  avoid 
surface  oscillations,  and  thereby  obtain 
more  precise  readings,  the  hook  gage 
should  be  set  up  in  a  still  box  com- 
municating with  the  channel.  The 
channel  end  of  the  opening  or  pipe 
leading  into  the  still  box  must  be  flush 
with  a  flat  surface  set  parallel  to  the 
direction  of  flow,  and  the  pipe  itself 
must  be  normal  to  this  direction. 

The  channel  end  of  the  pipe  must  be 
set  far  enough  above  the  weir  to  avoid 
the  slope  of  the  surface  curve,  but  not 
so  far  as  to  increase  the  head  by  the 
natural  slope  of  the  stream. 

If  the  formula  of  any  particular  ex- 
perimenter is  to  be  used,  his  location 
for  the  still  box  should  be  duplicated. 

Proportioning  Weirs. — To  illustrate 
the  method  of  proportioning  a  weir, 
suppose  that  the  stream  to  be  measured 
is  5-1/2  ft.  wide  and  1-1/2  ft.  deep, 
and  that  its  average  velocity,  deter- 
mined by  timing  a  float  over  a  meas- 
ured distance  or  by  using  a  current 
meter  or  a  Pitot  tube  (Arts.  26  and 
27),  is  approximately  4  ft.  per  second. 

The  flow  is  then  approximately  1980  cu.  ft.  per  minute.  To 
determine  the  size  of  weir  which  will  flow  approximately  this 
amount,  try  first  a  depth  of  say  10  in.  From  Table  17  it  is 
found  that  each  inch  of  length  for  this  depth  will  deliver 
12.64  cu.  ft.  per  minute.  The  required  length  of  weir  would 


FIG.  50. — Hook  gage. 


56  ELEMENTS  OF  HYDRAULICS 

then  be   T^QA  =  156.6,  which  is  fifteen   and  two-thirds  times 

the  depth  and  therefore  too  long  by  Rule  2  of  the  specifications. 

Since  the  weir  must  evidently  be  deeper,  try  18  in.      From 

the  table  the  discharge  per  linear  inch  for  this  depth  is  30.54 

cu.  ft.  per  minute,  and  consequently  the  required  length  would  be 

1980 

3Q54    =  64.8  in.,  which  is  now  only  3.6  times  the  depth  and 

therefore  too  short. 

By  further  trial  it  is  found  that  a  depth  of  15  in.  gives  a 

1980 
length  of  23~23  =  ^5'2  *n*'   wm^h  *s  5-^  times    the  depth  and 

therefore  comes  within  the  limits  required  by  Rule  2. 

Suppose  then  that  the  notch  is  made  7  ft.  long  and  say  20 
in.  deep,  so  that  the  depth  may  be  increased  over  the  calcu- 
lated amount  if  necessary.  If  then  the  width  of  the  pond  back 
of  the  weir  is  not  50  per  cent,  greater  than  the  width  of  the  notch, 
or  if  the  velocity  of  flow  should  be  in  excess  of  1  ft.  per  second, 
the  pond  should,  if  possible,  be  enlarged  or  deepened  to  give  the 
desired  result.  With  the  weir  so  constructed  suppose  that  the 
depth  of  water  over  the  stake  back  of  the  weir  is  found  to  be 
15-1/8  in.  From  the  table  the  discharge  per  linear  inch  cor- 
responding to  this  head  is  found  to  be  23.52  cu.  ft.  per  minute, 
and  this  multiplied  by  84,  the  length  of  the  weir  in  inches,  gives 
1975.7  cu.  ft.  per  minute  for  the  actual  measured  discharge. 

12.  TIME  REQUIRED  FOR  FILLING  AND  EMPTYING  TANKS 

Change  in  Level  under  Constant  Head. — To  find  the  time 
required  to  raise  or  lower  the  water  level  in  a  tank,  reservoir,  or 
lock,  let  A  denote  the  area  of  the  orifice  through  which  the  flow 
takes  place  and  K  its  coefficient  of  discharge  or  efflux.  Several 
simple  cases  will  be  considered. 

The  simplest  case  is  that  in  which  the  water  level  in  a  tank  is 
raised,  say  from  AB  to  CD  (Fig.  51),  by  water  flowing  in  under  a 
constant  head  h.  Let  V  denote  the  total  volume  of  water  flow- 
ing in,  represented  in  cross  section  by  the  area  ABCD  in  the  figure. 
Then  since  the  discharge  Q  through  the  orifice  per  second  is 

Q  =  KA  V20&, 
the  time  t  in  seconds  required  to  raise  the  surface  to  the  level  CD  is 


HYDROKINETICS 


57 


Varying  Head. — It  is  often  necessary  to  find  the  time  required 
to  empty  a  tank  or  reservoir,  or  raise  or  lower  its  level  a  certain 
amount.  A  common  case  is  that  in  which  the  level  is  to  be  raised 
or  lowered  from  AB  to  CD  (Fig.  52)  by  flow  through  a  sub- 
merged orifice,  the  head  on  one  side,  EF,  of  the  orifice  being 
constant.  If  the  cross  section  of  the  tank  is  variable,  let  Y  denote 
its  area  at  any  section  mn.  In  the  time  dt  the  level  changes  from 
the  height  y  to  y  —  dy,  and  consequently  the  volume  changes  by 
the  amount 

dV  =  Ydy. 

But  by  considering  the  flow  through  the  orifice,  of  area  At  the 
volume  of  flow  in  the  time  dt  is 


FIG.  51. 


dV 

Hence,  by  equating  these  values  of  dV,  we  obtain  the  relation 

dt  =  Ydy 


whence 


t  = 


KA 


(32) 


Canal  Lock. — A  practical   application  of  Eq.  (32)  is  in  find- 
ing the  time  required  to  fill  or  empty  a  canal  lock.     For  an 


58 


ELEMENTS  OF  HYDRAULICS 


ordinary  rectangular  lock  of  breadth  6  and  length  I,  the  cross  sec- 
tion is  constant,  namely  Y  =  bl,  and  consequently  the  expression 
for  the  time  integrates  into 


t  = 


bl 


KAV2g!      Vy        KAVg 


-  (  VH  -  Vh). 


(33) 


Rise  and  Fall  in  Connected  Tanks. — When  one  tank  discharges 
into  another  without  any  additional  supply  from  outside,  the 


Level  Raised 


T 


Level  Lowered 
FIG.  52. 


level  in  one  tank  falls  as  that  in  the  other  rises.  If  both  tanks  are 
of  constant  cross  section,  then  when  the  level  in  one  tank  has  been 
lowered  a  distance  y,  that  in  the  other  tank  will  have  been  raised 
a  distance  y'  (Fig.  53),  such  that  if  M  and  N  denote  their  sec- 
tional areas,  respectively, 

My  =  Ny'. 


HYDROKINETICS 


59 


In  the  interval  of  time  dt  suppose  y  changes  to  y  +  dy.  Then  con- 
sidering the  flow  through  the  orifice  of  area  A,  as  in  the  preceding 
case,  we  have 

Mdy  =  KA^2g[H  -  (y  +  yf)]dt, 

or,  since  y'  =  -TT-,  this  may  be  written 

MdyilN 


dt  = 


-  y(M 


Simplifying  this  expression  and  integrating,  the  resulting  expres- 
sion for  the  time  t  is  found  to  be 


r 
1. 


FIG.  53. 


KA 


r 

Jo 


dy 


+  N) 


KA-iYg 


-  y(M  +  N) 


]• 

-I  o 

Substituting  the  given  limits,  the  time  t  required  to  lower  the  level 
a  distance  D  is 

VNH  -  VNH  -  D(M  +  N)  I    (34) 


When  the  level  becomes  the  same  in  both  tanks,  since  the 
volume  discharged  by  one  is  received  by  the  other,  we  have 


or 


MD  =  N(H  -  D), 
NH 


D 


M  +  N 


60 


ELEMENTS  OF  HYDRAULICS 


Substituting  this  value  of  D  in  Eq.  (34),  it  becomes 

2MNVH 

KAV2g(M+  N)' 


t  = 


(35) 


which  is  therefore  the  length  of  time  required  for  the  water  in  the 
tanks  to  reach  a  common  level. 

Marietta's  Flask. — It  is  sometimes  desirable  in  measuring  flow 
to  keep  the  head  constant.  It  is  difficult  to  accomplish  this  by 
keeping  the  supply  constant,  a  more  convenient  method  being 
by  the  arrangement  shown  in  Fig.  54,  which  is  known  as  Mariotte's 
Flask.  This  consists  of  putting  an  air-tight  cover  on  the  tank, 
having  a  corked  orifice  holding  a  vertical  pipe  open  to  the  atmos- 


FIG.  54. 

phere.  Since  the  pressure  at  the  lower  end  A  of  the  tube  is 
always  atmospheric,  the  flow  is  the  same  as  though  the  water 
level  was  constantly  maintained  at  this  height.  Therefore  as 
long  as  the  water  level  does  not  sink  below  the  bottom  of  the  pipe, 
the  effective  head  on  the  orifice  is  its  distance  h  below  the  bottom 
of  the  pipe,  and  the  discharge  is  given  by  the  formula 


Q  =  KAV2gh. 


(36) 


13.  FLOW  THROUGH  SHORT  TUBES  AND  NOZZLES 


Standard  Mouthpiece. — When  a  short  tube  (adjutage,  mouth- 
piece or  nozzle)  is  added  to  an  orifice,  the  flow  through  the  open- 
ing is  changed  both  in  velocity  and  in  amount.  In  general  the 
velocity  is  diminished  by  the  mouthpiece,  due  to  increased  fric- 


HYDROKINETICS  61 

tional  resistance,  whereas  the  quantity  discharged  may  be  either 
increased  or  diminished,  depending  on  the  form  of  the  mouth- 
piece. 

What  is  called  the  standard  mouthpiece  consists  of  a  circular 
tube  projecting  outward  from  a  circular  orifice,  and  of  length 
equal  to  two  or  three  diameters  of  the  orifice  (Fig.  56).  At  the 
inner  end  of  the  tube  the  jet  is  contracted  as  in  the  case  of  a 
standard  orifice,  but  farther  out  it  expands  and  fills  the  tube. 
The  velocity  of  the  jet  is  reduced  by  this  form  of  mouthpiece  to 

v  =  0.82  V2J/S, 

which  is  considerably  less  than  for  a  standard  orifice,  but  since 
there  is  no  contraction,  the  quantity  discharged  is 

Q  =  0.824  V2p, 

where  A  denotes  the  area  of  the  orifice.  The  discharge  is  there- 
fore nearly  one-third  larger  than  for  a  standard  orifice  of  the  same 
area  with  complete  contraction  (Fig.  55). 

,  Stream  -line  Mouthpiece.  —  By  rounding  the  inner  edge  of  the 
mouthpiece  so  that  its  contour  approximates  the  form  of  a  stream 
line,  the  velocity  of  the  jet  is  greatly  increased,  its  value  for  the 
relative  dimensions  shown  in  Fig.  57  being  about 


v  =  0.96 

and  since  the  jet  suffers  no  contraction,  the  quantity  discharged  is 

Q  =  0.964 


the  area  A,  as  before,  referring  to  the  area  of  the  orifice. 

Borda  Mouthpiece.  —  A  mouthpiece  projecting  inward  and 
having  a  length  of  only  half  a  diameter  is  called  a  Borda  mouth- 
piece (Fig.  58)  .  The  velocity  is  greatly  increased  by  this  form  of 
mouthpiece,  its  value  being  about 

v  =  0.99  V2p, 

but  the  contraction  of  the  jet  is  more  than  for  a  standard  orifice, 
so  that  the  discharge  is  only 

Q  =  0.534 
where  A  denotes  the  area  of  the  orifice. 


62 


ELEMENTS  OF  HYDRAULICS 


FIG.  57. 


FIG.  58. 


Conical  Diverging 
Tube 


Re-entrant  Tube  :":::f-- 


^ 


F!G.    59. 


FIG.  60. 


Venturi  Adjutage 
Angle  a  =  5°  to  8° 

Q=1.5  A\/2gh 
Area  A  Measured  on 
A   Section  AB 


•  _        ^ii  1 1 1  - 


Conical  Converging 

Tubeo 
Angle  a=5°to  10° 


FIG.  61. 


FIG.  62. 


HYDROKINETICS 


63 


Fire  Hose;  Smooth  Cone  Nozzle 
Q=.WA~V  2gh 

FIG.  63. 


Fire  Hose;  Smooth  Convex  Nozzle 
Q=.Q7A\/ 

FIG.  64. 


Fire  Hose;  Square  Ring  Nozzle 
A\rZrtT 

FIG.  65. 


Fire  Hose;  Undercut  Ring  Nozzle 
Q=.1lAV2 

FIG.  66. 


If,  however,  the  length  of  the  mouthpiece  is  increased  to  two  or 
three  diameters  (Fig.  59)  the  discharge  is  increased  nearly  50  per 
cent.,  becoming 

Q  =  0.724  ^J2gh. 

Diverging  Conical  Mouthpiece. — For  a  conical  diverging  tube 
with  sharp  edge  at  entrance  (Fig.  60)  the  jet  contracts  at  the 
inner  end  as  for  an  orifice,  but  farther  on  expands  so  as  to  fill  the 
tube  at  outlet  provided  the  angle  of  divergence  is  not  over  8°. 
The  discharge  is  therefore  greater  than  for  a  standard  mouth- 
piece, its  amount  referred  to  the  area  A  at  the  smallest  section 
being 

Q  =  0.95  A  V20S. 

Venturi  Adjutage. — If  the  entrance  to  a  diverging  conical 
mouthpiece  has  a  stream-line  contour,  it  is  called  a  Venturi 
adjutage  (Fig.  61).  In  experiments  by  Venturi  and  Eytelwein 
with  diverging  mouthpieces  of  the  relative  dimensions  shown  in 
Fig.  61,  a  discharge  was  obtained  nearly  two  and  one-half  times 
as  great  as  for  a  standard  orifice  of  the  same  diameter  as  that  at 
the  smallest  section,  or  about  twice  that  for  a  standard  short  tube 


64  ELEMENTS  OF  HYDRAULICS 

of  this  diameter,  the  formula  for  discharge  referred  to  the  area  A 
at  the  smallest  section  being 

Q  =  1.55AV20S. 

Converging  Conical  Mouthpiece. — In  the  case  of  a  conical 
converging  tube  with  sharp  corners  at  entrance  (Fig.  62)  the  jet 
contracts  on  entering  and  then  expands  again  until  it.  fills  the  tube, 
the  most  contracted  section  being  just  beyond  the  tip,  and  the 
greatest  discharge  occurring  for  an  angle  of  convergence  of  approxi- 
mately 13°. 

Fire  Nozzles. — The  fire  nozzles  shown  in  Figs.  63,  64,  65  and 
66  are  practical  examples  of  converging  mouthpieces.  The 
smooth  cone  nozzle  with  gradually  tapering  bore  has  been  found 
to  be  the  most  efficient,  the  coefficient  of  discharge  for  the  best 
specimen  being  0.977  with  an  average  coefficient  for  this  type 
of  0.97.  For  a  square  ring  nozzle  like  that  shown  in  Fig.  65  the 
coefficient  of  discharge  is  0.74;  and  for  the  undercut  type  shown  in 
Fig.  66  the  coefficient  of  discharge  is  0.71. 

14.  KINETIC  PRESSURE  IN  A  FLOWING  LIQUID 

Kinetic  Pressure. — For  a  liquid  at  rest,  the  normal  pressure 
exerted  by  it  on  any  bounding  surface  is  called  the  hydrostatic 
pressure  and  is  given  by  the  expression  deduced  in  Art.  2,  namely, 

p  =  p'  +  yh. 

If  a  liquid  is  in  motion,  however,  the  normal  pressure  it  exerts 
on  the  walls  of  the  vessel  containing  it,  or  on  the  bounding 
surface  of  a  liquid  vein  or  filament,  follows  an  entirely  different 
law,  as  shown  below. 

To  distinguish  the  hydrostatic  pressure  from  the  normal  pres- 
sure exerted  on  any  bounding  surface  by  a  liquid  in  motion,  the 
latter  will  be  called  the  kinetic  pressure. 

Bernoulli's  Theorem. — To  determine  the  kinetic  pressure  at 
any  point  in  a  flowing  liquid,  consider  a  small  tube  or  vein  of  the 
liquid  bounded  by  stream  lines,  as  explained  in  Art.  8,  and  follow 
the  motion  of  the  liquid  through  this  tube  for  a  brief  interval  of 
time. 

Let  A  and  A'  denote  the  areas  of  two  normal  cross  sections  of 
the  vein  (Fig.  67).  Then  since  the  liquid  is  assumed  to  be  in- 
compressible, the  volume  Ad  displaced  at  one  end  of  the  tube 


HYDROKINETICS 


65 


must  equal  the  volume  A'd!  displaced  at  the  other  end.  If  p 
denotes  the  average  unit  pressure  on  A,  and  p'  on  A',  the  work 
done  by  the  pressure  on  the  upper  cap,  A,  is 

+  pAd, 

and  that  on^the  lower  cap,  A',  is 

-  p'A'd'. 

Also,  if  h  denotes  their  difference  in  static  head,  as  indicated  in 
Fig.  67,  the  work  done  by  gravity 
in  the  displacement  of  the  vol- 
ume Ad  a  distance  h  is  IP 


"T 


vAdh. 

Since  the  forces  acting  on  the 

lateral  surface  of  the  vein  are 

normal  to  this  surface  they  do 

no  work.     Assuming,  then,  the 

case  of  steady  flow,  that  is  to 

say,  assuming  that  each  particle 

arriving  at  a  given  cross  section 

experiences  the  same  velocity  and 

pressure  as  that  experienced  by 

the   preceding   particle   at   this 

point,  so  that  the  velocities  v  and  v'  through  the  caps  A  and  A' 

are  constant,  the  change  in  kinetic  energy  between  these  two 

positions  is 


FIG.  67. 


Therefore,  equating  the  total  work  done  to  the  change  in  energy, 
the  result  is 


pAd  -  p'A'd'  +  7Adh  = 
or,  since  Ad  =  Afdf,  this  reduces  to 


v"*  -  z;2), 


P' 


yy^2 
2g 


TV 


(37) 


This  result  is  known  as  Bernoulli's  Theorem,  and  shows  that  in 
the  case  of  steady  parallel  flow  of  an  ideal  liquid,  an  increase  in 
velocity  at  any  point  is  accompanied  by  a  corresponding  decrease 

5 


66 


ELEMENTS  OF  HYDRAULICS 


in  kinetic  pressure,  or  vice  versa,  in  accordance  with  the  relation 
just  obtained. 

Kinetic  Pressure  Head. — If  the  theoretic  heads  corresponding 
to  the  velocities  v  and  v'  are  denoted  by  H  and  H',  respectively, 
then  in  accordance  with  Torricelli's  theorem  (Art.  8)  we  have 

v2  v'2 

•    ^  TT      •  TTf      _ . 

2</ '  2g' 

and  consequently  Eq.  (37)  may  be  written  in  the  form 

P'  =  P  +  7(h  +  H  -  H'),  (38) 

which  is  a  convenient  form  from  which  to  compute  the  kinetic 
pressure  at  any  given  point. 

If  this  relation  is  written  in  the  form 

—  +  H'  =  -  +  H  +  h,  (39) 

then  since   p/y  is   the   head  corresponding  to  the  hydrostatic 
pressure  p,  each  term  is  a  length,  and  Bernoulli's  theorem  may  be 

expressed  by  saying  that: 

In  the  case  of  steady,  par- 
allel flow  of  an  ideal  liquid, 
the  sum  of  the  pressure  head, 
velocity  head   and  potential 
head  is  a  constant  quantity 
ZLZ    for  any  particle  throughout 
^^    its  course. 

Application  to  Standard 
Mouthpiece. — An  illustra- 
tion of  Bernoulli's  theorem 
is  afforded  by  the  flow 
through  a  standard  mouth- 
piece. At  the  contracted 
section  A  (Fig.  68)  the  ve- 
locity is  evidently  greater 
than  at  the  outlet  B. 
Therefore,  by  Bernoulli's  theorem,  the  kinetic  pressure  must  be 
less  at  A  than  at  B.  Thus  if  a  piezometer  is  inserted  in  the 
mouthpiece  at  A,  the  liquid  in  it  will  rise,  showing  that  the 
pressure  in  the  jet  at  this  point  is  less  than  atmospheric.  It 
was  found  by  Venturi,  and  can  also  be  proved  theoretically,  that 
for  a  standard  mouthpiece  the  negative  pressure  head  at  A  is 


Hi 


± 


FIG.  68. 


HYDROKINETICS  67 

approximately  three-fourths  of  the  static  head  on  the  opening, 
or,  referring  to  Fig.  68, 

h,  =  lh. 

16.  VENTURI  METER 

Principle  of  Operation. — The  Venturi  meter,  invented  by 
Clemens  Herschel  in  1887  for  measuring  flow  in  pipe  lines, 
illustrates  an  important  commercial  application  of  Bernoulli's 
theorem.  This  device  consists  simply  of  two  frustums  of  conical 
tubes  with  their  small  ends  connected  by  a  short  cylindrical 
section,  inserted  in  the  pipe  line  through  which  the  flow  is  to  be 
measured  (Fig.  69).  If  a  pressure  gage  is  inserted  in  the  pipe 


B 


FIG.  69. 

line  at  any  point  A  and  another  at  the  throat  of  the  meter  B, 
as  indicated  in  the  figure,  it  will  be  found  that  the  pressure  at 
B  is  less  than  at  A. 

Formula  for  Flow. — Let  VA  and  VB  denote  the  velocities  at  A 
andB,  and  pA  and  pB  the  kinetic  pressures  at  these  points,  respect- 
ively. Then  since  both  points  are  under  the  same  static  head, 
Bernoulli's  theorem,  disregarding  frictional  losses,  gives  the 
relation 

VA2    ,PA_     VB*    ,    Pji 
2g  ~~  y          2g   "  j' 

If  a  and  b  denote  the  cross-sectional  areas  at  A  and  B,  the  dis- 
charge Q  is  given  by 

Q  =  avA  =  bvB 
whence 

Q  Q 

VA  =  a'>    VB  =  b' 

Substituting  these  values  of  VA  and  VB  in  the  preceding  equation 
and  solving  for  Q,  the  result  is 


n  ab 

Q  = 


68 


ELEMENTS  OF  HYDRAULICS 


If  hA  and  hB  denote  the  static  piezometer  heads  corresponding  to 
the  kinetic  pressures  pA  and  PB,  respectively,  this  formula  may 
be  written 

ab 

(40) 


FIG.  70. — Venturi  meter  and  recording  gage,  manufactured  by  the  Builders 

Iron  Foundry. 

Ordinarily  the  throat  diameter  in  this  type  of  meter  is  made 
one-third  the  diameter  of  the  main  pipe,  in  which  case  a  =  96. 


HYDROKINETICS 


69 


70  ELEMENTS  OF  HYDRAULICS 

If  then,  h  denotes  the  difference  in  piezometer  head  between  the 
up-stream  end  and  the  throat,  the  formula  for  discharge,  ignoring 
frictional  losses,  becomes 

Q  =  1.0062  b\/2gh.  (41) 

By  experiment  it  has  been  found  that  ordinarily  for  all  sizes  of 
Venturi  meters  and  actual  velocities  through  them,  the  actual 
discharge  through  the  meter  is  given  by  the  empirical  formula 

Q  =  (0.97  ±  o.03)b\4gh.  (42) 

Commercial  Meter. — A  typical  arrangement  of  meter  tube  and 
recording  apparatus  is  shown  in  Fig.  70,  the  lower  dial  indicating 
the  rate  of  flow,  and  the  upper  dial  making  a  continuous  auto- 
graphic record  of  this  rate  on  a  circular  chart. 

Catskill  Aqueduct  Meter. — The  Venturi  meter  affords  the  most 
accurate  method  yet  devised  for  measuring  the  flow  in  pipe  lines. 
Fig.  71  shows  one  of  the  three  large  Venturi  meters  built  on 


FIG.  72. — Venturi  rate  of  flow  controller  manufactured  by  the  Simplex 
Valve  and  Meter  Co. 

the  line  of  the  Catskill  Aqueduct,  which  is  part  of  the  water 
supply  system  of  the  City  of  New  York.  Each  of  these  meters  is 
410  ft.  long  and  is  built  entirely  of  reinforced  concrete  except  for 
the  throat  castings  and  piezometer  ring,  which  are  of  cast  bronze. 
Provision  is  also  made  in  connection  with  the  City  aqueduct  for 
the  installation  of  a  Venturi  meter  upon  each  connection  between 
the  aqueduct  and  the  street  distribution  pipes. 

Rate  of  Flow  Controller. — Fig.  72  illustrates  a  rate  of  flow  con- 
troller operated  by  the  difference  in  pressure  in  a  Venturi  tube. 


HYDROKINETICS  71 

This  apparatus  is  designed  for  use  in  a  water  pipe  or  conduit 
through  which  a  constant  discharge  must  be  maintained  regard- 
less of  the  head  on  the  valve.  It  consists  of  a  perfectly  balanced 
valve  operated  by  a  diaphragm  which  is  actuated  by  the  differ- 
ence in  pressure  between  the  full  and  contracted  sections  of  a 
Venturi  tube.  The  valve  and  diaphragm  are  balanced  by  an 
adjustable  counter- weight,  which  when  set  for  any  required  rate  of 
flow  will  hold  the  valve  discs  in  the  proper  position  for  that  flow. 

16.  FLOW  OF  WATER  IN  PIPES 

Critical  Velocity. — Innumerable  experiments  and  investiga- 
tions have  been  made  to  determine  the  laws  governing  the  flow  of 
water  in  pipes,  but  so  far  with  only  partial  success,  as  no  general 
and  universal  law  has  yet  been  discovered. 

Experiments  made  by  Professor  Osborne  Reynolds  have  shown 
that  for  a  pipe  of  a  given  diameter  there  is  a  certain  critical 
velocity,  such  that  if  the  velocity  of  flow  is  less  than  this  critical 
value,  the  flow  proceeds  in  parallel  filaments  with  true  stream- 
line motion;  whereas  if  this  critical  value  is  exceeded,  the  flow 
becomes  turbulent,  that  is,  broken  by  whirls  and  eddies  and 
similar  disturbances.  The  results  of  Professor  Reynold's  experi- 
ments showed  that  at  a  temperature  of  60°  F.  this  critical  velocity 
occurred  when 

Dva  =  0.02 

where  D  denotes  the  diameter  of  the  pipe  in  feet  and  va  is  the 
average  velocity  of  flow  in  feet  per  second. 

For  parallel,  or  non-sinuous,  flow  it  is  possible  to  give  a  theoret- 
ical explanation  of  what  occurs  and  deduce  the  mathematical  law 
governing  it,  as  shown  below.  No  one,  however,  has  yet  ex- 
plained why  the  flow  suddenly  becomes  turbulent  at  the  critical 
velocity,  or  what  law  governs  it  subsequently. 

Viscosity  Coefficient. — The  loss  of  energy  accompanying  pipe 
flow  is  due  to  the  internal  resistance  arising  from  the  viscosity 
of  the  liquid.  This  shear  or  drag  between  adjacent  filaments  is 
analogous  to  ordinary  friction  but  follows  entirely  different  laws. 
Unlike  friction  between  the  surfaces  of  solids,  fluid  friction  has 
been  found  by  experiment  to  be  dependent  on  the  temperature 
and  the  nature  of  the  liquid;  independent  of  the  pressure;  and, 
for  ordinary  velocities  at  least,  approximately  proportional  to  the 


72 


ELEMENTS  OF  HYDRAULICS 


difference  in  velocity  between  adjacent  filaments.  When  this 
difference  in  velocity  disappears,  the  frictional  resistance  also 
disappears. 

The  constant  of  proportionality  required  to  give  a  definite 
numerical  value  to  fluid  friction  is  called  the  viscosity  coefficient 
and  will  be  denoted  by  ju.  This  coefficient  M  is  an  empirical  con- 
stant determined  by  experiment,  the  values  tabulated  below  being 
the  result  of  experiments  made  by  O.  E.  Meyer. 


Temperature     in     degrees 
Fahr. 

50° 

60° 

65° 

70°' 

Viscosity    coefficient   //    in 
Ib.  sec. 

32  X  10~6 

28  X  10-6 

26  X  10~6 

24  X  10~6 

ft.2 

FlG   73 


The  dimensions  of  /*  are,  of  course,  such  as  to  make  the  equa- 
tion in  which  it  appears  homogeneous  in  the^units  involved,  as 
will  appear  in  what  follows. 

Parallel  (non-sinuous)  Flow.1 — Consider  non-sinuous  flow  in  a 
straight  pipe  of  uniform  circular  cross  section,  that  is,  at  a  veloc- 
ity less  than  the  critical  ve- 
locity and  therefore  such 
that  the  filaments  or  stream 
lines  are  all  parallel  to  the 
axis  of  the  pipe.  By  reason 
of  symmetry  the  velocity 
of  any  particle  depends 
only  on  its  distance  from 

the  axis  of  the  pipe.  Let  v  denote  the  velocity  of  any  particle 
and  x  its  distance  from  the  center  (Fig.  73) .  Then  if  x  changes 
by  an  amount  dx,  the  velocity  changes  by  a  corresponding 
amount  dv,  and  since  the  velocity  is  least  near  the  pipe  walls,  v 

decreases  as  x  increases  and  consequently  the  ratio  -r-  is  negative. 

Since  the  pipe  is  assumed  to  be  of  constant  cross  section  and  the 
flow  uniform  and  parallel,  the  forces  acting  on  any  element  of 
volume  must  be  in  equilibrium.  Considering  therefore  a  small 
water  cylinder  of  radius  x  and  length  dyr  in  order  to  equilibrate 
the  frictional  resistance  acting  on  the  convex  surface  of  this 
cylinder  there  must  be  a  difference  in  pressure  on  its  ends. 

1  The  following  derivation  is  substantially  that  given  by  Foppl  in  his 
Dynamik. 


HYDROKINETICS  73 

This  explains  the  fall  in  pressure  along  a  pipe,  well  known  by 
experiment. 

Let  dp  denote  the  difference  in  pressure  in  a  length  dy.     Then 
the  difference  in  pressure  on  the  ends  of  a  cylinder  of  radius  x  is 

(•7rx2)dp  and  the  shear  on  its  convex  surface  is  (27rxdy)  JJL-T-  •     Equat- 

ing these  two  forces  and  remembering  that  -y-  is  negative,  we 
obtain  the  relation 


Also,  since  the  difference  in  pressure  on  the  ends  of  any  cylinder 
is  proportional  to  its  length,  we  have 

dp  _  pi  -p2  _ 
dy~         I 

where  pi  and  p2  denote  the  unit  pressures  at  two  sections  at  a 
distance  I  apart,  and  the  constant  ratio  is  denoted  by  k  for 
convenience. 

Substituting  dp  =  kdy  from  the  second  equation  in  the  first  and 
cancelling  common  factors,  we  have  finally 

dv  k 

dx  =     ~  2ju  x' 
whence,  by  integration, 

kx2 
'  4^  +  c' 

where  c  denotes  a  constant  of  integration. 

To  determine  c  assume  that  the  frictional  resistance  between 
the  pipe  wall  and  the  liquid  follows  the  same  law  as  that  be- 
tween adjacent  filaments  of  the  liquid.  Then  it  follows  that 
the  liquid  in  contact  with  the  pipe  must  have  zero  velocity,  as 
otherwise  it  would  experience  an  infinite  resistance.  This  seems 
also  to  be  confirmed  by  the  experiments  of  Professor  Hele-Shaw, 
who  showed  that  in  the  case  of  turbulent  flow  there  was  always  a 
thin  film  of  liquid  adjoining  the  pipe  walls  which  showed  true 
stream-line  motion,  proving  that  its  velocity  was  certainly  less 
than  the  critical  velocity  and  therefore  small.  Furthermore,  the 
walls  of  commercial  pipes  are  comparatively  rough  and  conse- 
quently a  thin  skin  or  layer  of  liquid  must  be  caught  in  these 
roughnesses  and  held  practically  stationary. 

Assuming  then  that  v  =  0  when  x  =  r,  and  substituting  this 


74  ELEMENTS  OF  HYDRAULICS 

pair  of  simultaneous  values  in  the  above  equation,  the  value  of  c 
is  found  to  be 

fcr2 


and  consequently 


(r»  -  x").  (43) 


This  is  the  equation  of  a  parabola,  and  therefore  the  velocity  dia- 
gram is  a  parabolic  arc  with  its  vertex  in  the  axis  of  the  pipe; 
that  is,  the  velocity  is  a  maximum  at  the  center  where  x  =  0,  its 
value  being 

fcr2 

V 


Vmax    = 


Average  Velocity  of  Flow  in  Small  Pipes. — Let  the  discharge 
through  any  cross  section  of  the  pipe  be  denoted  by  Q.  Then  if 
the  velocity  at  any  radius  x  is  denoted  by  v,  we  have 


-f 


Q  =  I    2irxdxv, 

)o 

k 
or,  since  v  =  j-  (r2  —  z2), 

this  becomes 


But  if  va  denotes  the  average  velocity  of  flow  we  also  have 

Q  =  tfa&rr2) 
whence  by  substituting  the  above  value  for  Q,  we  have 


Q 


Comparing  this  expression  with  that  previously  obtained  for  the 
maximum  velocity,  it  is  evident  that  the  maximum  velocity  is 
twice  the  average  velocity  of  flow. 

Loss  of  Head  in  Small  Pipes.  —  The  loss  in  pressure  in  a  length 
I  is  given  by  the  relation  obtained  above,  namely, 


HYDROKINETICS  75 

or,  if  the  difference  in  head  corresponding  to  this  difference  in 
pressure  is  denoted  by  h,  then,  since  p  =  yh,  we  have 

i       •     u     j    z,       PI  -  P*       kl 
loss  in  head,  h  =  -          -  =  — 


Substituting  in  this  relation  the  value  of  k  in  terms  of  the  average 
velocity  of  flow,  the  result  is 

8Mlva  _  2Mly_0. 
~  kr2T     "  kd2T 

For  small  pipes,  therefore,  the  loss  of  head  is  proportional  to  the 
first  power  of  the  average  velocity,  and  inversely  proportional 
to  the  square  of  the  diameter  of  the  pipe. 

This  result  has  been  verified  experimentally  for  small  pipes  by 
the  experiments  carried  out  by  Poiseuille. 

Ordinary  Pipe  Flow.  —  Under  the  conditions  usually  found  in 
practice  the  velocity  of  flow  exceeds  the  critical  velocity  and  con- 
sequently the  flow  is  turbulent  and  a  greater  amount  of  energy  is 
dissipated  in  overcoming  internal  resistance  than  in  the  case  of 
parallel  flow.  The  result  of  Professor  Reynold's  experiments 
indicated  that  the  loss  of  head  in  turbulent  flow  was  given  by  the 
relation 


In  commercial  pipes  the  degree  of  roughness  is  a  variable  and  un- 
certain quantity,  so  that  the  exact  loss  of  head  cannot  be  pre- 
dicted with  accuracy.  Practical  experiments  have  shown,  how- 
ever, that  ordinarily  the  loss  in  head  is  proportional  to  the  square 
of  the  average  velocity,  so  that  the  relation  becomes 


Since  the  theoretical  head  corresponding  to  a  velocity  v  is 

v2 
A  =  ?r>  the  expression  for  the  loss  in  head  for  a  circular  pipe 

*Q 
running  full  may  in  general  be  written 


or,  denoting  the  constant  of  proportionality  by/,  this  becomes 


76  ELEMENTS  OF  HYDRAULICS 

Here/  is  an  empirical  constant,  depending  on  the  condition  of  the 
inner  surface  of  the  pipe,  and  is  determined  by  experiment. 
Eq.  (45)  is  identical  with  Chezy's  well-known  formula 


as  will  be  shown  in  Art.  19. 

17.  PRACTICAL  FORMULAS  FOR  LOSS  OF  HEAD  IN  PIPE  FLOW 

Effective  and  Lost  Head.  —  In  the  case  of  steady  flow  through 
long  pipes,  much  of  the  available  pressure  head  disappears  in 
frictional  and  other  losses,  so  that  the  velocity  is  greatly  dimin- 
ished. Thus  if  h  denotes  the  static  head  at  the  outlet  and  hi 
the  head  lost  in  overcoming  frictional  and  other  resistances  to 
flow,  the  velocity  v  at  the  outlet  is  given  by  the  relation 


or  its  equivalent, 

h  =  5  +  h"  (46) 

The  lost  head  hi  is  the  sum  of  a  number  of  terms,  which  will  be 
considered  separately. 

Loss  at  Entrance.  —  A  certain  amount  of  head  is  lost  at  the 
entrance  to  the  pipe,  as  in  the  case  of  a  standard  adjutage.  If 
v  denotes  the  velocity  due  to  the  head  h  with  no  losses,  then 

v* 

•     .  ^ 

whereas  if  VA  denotes  the  actual  velocity  of  flow  the  head  corre- 
sponding to  this  velocity  is 

v  -  v^- 

20 
The  head,  hi,  lost  at  entrance,  is  therefore 


If  Cv  denotes  the  velocity  coefficient  for  the  entrance,  then 

VA=  Cvv, 


HYDROKINETICS  77 

and  consequently  the  expression  for  the  head  lost  at  entrance 
may  be  written 


=          _ 
~ 


2g        2g  2g         2g 


2g    \(V 

For  the  standard  short  tube  Cv  =  0.82  (Art.  13)  and  therefore 

1  1 

C^2  ~  ^  =  (0  82) 2  ~~  ^  =  '  ^e  nea^  l°s^  a^  enhance  is 
therefore 

v2 
hi  =  0.5  — •  (47) 

If  the  pipe  projects  into  the  reservoir,  Cv  —  0.72  (Art.  13),  and 
the  head  lost  at  entrance  is  thereby  increased  to 

hi  =  0.93  |~- 

For  ordinary  service  taps  on  water  mains  it  may  be  assumed  as 

hi  =  0.62 ~  • 

Friction  Loss. — In  flow  through  long  pipes  the  greatest  loss 
in  head  is  that  due  to  the  friction  between  the  liquid  and  the 
walls  of  the  pipe.  Let  d  denote  the  internal  diameter  of  the 
pipe  and  I  its  length.  Then  it  has  been  found  by  experiment 
(Art.  16)  that  the  head  lost  in  internal  friction,  or  friction  head 
as  it  is  called,  is  given  by  the  formula 

h'='fa'$'  (*8) 

the  quantity/  being  an  empirical  constant  determined  by  experi- 
ment. The  average  values  of  this  coefficient  for  cast  iron  pipes 
are 

For  new  smooth  pipes,  f  =  0.024 ;  .    . 

For  old  rusty  pipes,        f  =  0.03. 

For  more  exact  values  of  this  coefficient,  refer  to  Table  12. 

Bends  and  Elbows. — Bends  and  elbows  in  a  pipe  also  greatly 
diminish  the  effective  head.  From  experiments  by  Weisbach 


78 


ELEMENTS  OF  HYDRAULICS 


it  has  been  found  that  the  lost  head  due  to  a  sharp  elbow  of 
angle  a  (Fig.  74)  is  given  by  the  formula 


(So) 


where  m  is  a  function  of  the  angle  a,  given  by  the  equation 
m  =  0.9457  sin2    -        +  2.047  sin4 


Values  of  m,  calculated  from  this  formula  for  various  values  of  the 
angle  a,  are  tabulated  as  follows: 


a  = 

20°  \      30°  \      40° 

50°        60° 

70° 

80°       90° 

m  = 

.046 

.073 

.139 

.234 

.364 

.533 

.740 

.984 

FIG.  74. 


FIG.  75. 


For  a  curved  elbow  of  radius  R  and  central  angle  a  (Fig.  75) 
the  lost  head  is  given  by  the  formula 


where  the  coefficient  n  has  the  value 

n  =  0.131  +0.163(^ 


Values  of  n  calculated  from  this  formula  for  various  values  of  the 

ratio  ^  are  tabulated  below  for  convenience  in  substitution. 
it 


d 
R~ 

.2 

.3 

.4 

.5 

.6 

.8 

1.0 

1.2 

1.25 

1.3 

1.4 

1.6 

1.8 

2.0 

n  = 

.1311.  1331.138 

.145i.  1581.  206|    .2941    .4401    .487|    .539!    .66l|    .977|l.40|l.98 

Enlargement  of  Section. — A  sudden  enlargement  in  the  cross 
section  of  a  pipe  decreases  the  velocity  of  flow  and  causes  a  loss 
of  head  due  to  eddying  in  the  corners,  etc.  (Fig.  76).  If  the 


HYDROKINETICS 


79 


velocity  is  decreased  by  the  enlargement  from  v\  to  vz,  it  has  been 
found  by  experiment,  and  can  also  be  proved  theoretically,  that 
the  head  lost  in  this  way  is  given  by  the  formula 

111  Ju-k^  (S') 

To  obtain  a  more  convenient  expression  for  h4,  let  a  denote  the 


%?. 


'  " 


FIG.  76. 

area  of  cross  section  of  the  smaller  pipe  and  A  of  the  larger. 
Then 

via  =  v2A, 
whence 


and  consequently  the  expression  for  h*  may  be  written 

V22/A  2 


(53) 


I - 


FIG.  77. 


Contraction  of  Section. — A  sudden  contraction  in  section  also 
causes  a  loss  in  head,  similar  to  that  due  to  a  standard  orifice  or 
adjutage  (Fig.  77).  The  lost  head  in  this  case  has  been  found  by 
experiment  to  be  given  by  the  equation 


(54) 


80 


ELEMENTS  OF  HYDRAULICS 


where  q  denotes  an  empirical  constant,  determined  experimentally. 
The  following  tabulated  values  of  the  coefficient  q  are  based  on 
experiments  by  Weisbach,  A  being  the  cross-sectional  area  of  the 
larger  pipe  and  a  of  the  smaller.1 


a 
I 

.1 

.2 

.3 

.4 

.5 

.6 

.7 

.8 

.9 

1.0 

q           |.362|.338|.308  1.267  |.22l|.164 

.  105  |.  053 

.015 

.000 

Gate  Valve  in  Circular  Pipe. — The  loss  in  head  due  to  a  partly 
closed  gate  valve  (Fig.  78)  has  been  determined  by  experiment 
for  different  ratios  of  height  of  opening  to  diameter  of  pipe  with 
the  following  results.2  In  this  Table,  x  denotes  the  height  of 

the  opening,  d  the  diameter  of  the  pipe,  h6  the  loss  in  head  and 

vz 
f  the  empirical  coefficient  in  the  formula  h&  =  f~-- 


X 

d 

i 

i 

4 

I 

\ 

1 

* 

7 
I 

i 

97.8 

17.0 

5.52     |      2.06 

0.81     |      0.26 

0.07 

FIG.  78. 


FIG.  79. 


FIG.  80. 


Cock  in  Circular  Pipe. — For  a  cock  in  a  cylindrical  pipe  (Fig. 
79)  the  coefficient  f  has  been  determined  in  terms  of  the  angle 
of  closure  0  with  the  following  results. 


e 

5°  I  10° 

15° 

20° 

25° 

30°  |  35° 

40° 

45°  |  50° 

55°  |  60° 

65° 

82° 

f 

.05 

.29 

.75 

1.56 

3.1 

5.47 

9.68 

17.3 

31.2 

52.6 

106 

206 

486 

Valve 
closed 

1  Hoskins,  Text-book  on  Hydraulics,  page  74. 

2  The  coefficient  for  losses  at  valves  are  based  on  experiments  by  Weisbach 
and  are  given  in  most  standard  texts  on   Hydraulics.      See  for  example 
Wittenbauer,  Aufgabensammlung,  Bd.  Ill,  S.  318;  Gibson,  Hydraulics  and 
its  Applications,  pp.  249,  250. 


HYDROKINETICS 


81 


Throttle  Valve  in  Circular  Pipe.  —  The  coefficient  {  in  the  for- 

v2 
mula/ie  =  ^  has  been  determined  experimentally  for  a  throttle 

valve  of  the  butterfly  type  (Fig.  80)  for  various  angles  of  closure 
with  results  as  follows: 


e 

5° 

10° 

20°  |  30°  |  40°   |  45°  |  50° 

60°  |  70° 

f 

0.24 

0.52 

1.54  |  3.91  |  10.80 

18.70 

32.6 

118  |  751 

Summary  of  Losses. — The  total  head,  hi,  lost  in  flow  through  a 
pipe  line  is  then  the  sum  of  the  six  partial  losses  in  head  mentioned 
above,  namely 

hi  =  hi  +  h2  +  hz  +  &4  +  hs  +  h6. 

The  values  of  these  six  terms  may  be  tabulated  as  follows: 


Loss  of  head  in  pipe  flow 

Head  lost  at  entrance 

*•  -  4 

Coefficient   modified   by    nature    of 
entrance  and  varies  from    .5  to   .9 

Friction  head 

*•-'/(£)       ' 

New  pipe,  /  =  .024 
Old  pipe,  /  =  .03 
See  table  12 

Head  lost  at  bends 
and  elbows 

fi3  =  m,n"    (Sharp  bend) 

*3  =  n  (9"d)£  (Curved 
elbow) 

TO  =    .9457  sin'  —  +  2-047  sin* 

/  d\3'5 
n  =    .131  +   .168(2) 

(f) 

Head  lost  at  sudden 
enlargement 

(»i  -  t>»>* 
*4  -    -*-*  — 

Head  lost  at  sudden 
contraction 

715  =  % 

For  values  of  coefficient,  see  Table 
page  80 

Head  lost  at  parti- 
ally closed  valve 

*'  =  4 

See  Tabular  values  of  f  ,    pag 
and  81 

es   80 

From  Eq.  (36)  we  have 

v2  v2 

h  =  T-  +  hi  =         -f- 


-f 


and  inserting  in  this  relation  the  values  tabulated  above  and  solv- 
ing for  the  velocity  of  flow,  v,  we  have 


2gh 


(55) 


82 


ELEMENTS  OF  HYDRAULICS 


Application. — To  give  a  simple  illustration  of  the  application 
of  Formula  (55),  suppose  it  is  required  to  find  the  velocity  of  flow 
fora  straight  new  cast  iron  pipe,  1  ft.  in  diameter  and  5000  ft.  long, 
with  no  valve  obstructions,  which  conducts  water  from  a  reservoir 
the  surface  of  which  is  150  ft.  above  the  outlet  of  the  pipe. 

In  this  case 


2gh 


2(32.2)150 


1+0.5+ 0.024 


=  8.9  ft.  per 


sec. 


and  the  discharge  is 

Q  =  Av  =  \  X  8.9  X  60  =  419.4  cu.  ft.  per  min. 


18.  HYDRAULIC  GRADIENT 

Kinetic  Pressure  Head. — In  the  case  of  steady  flow  through  a 
long  pipe,  if  open  piezometer  tubes  are  inserted  at  different 
points  of  its  length  and  at  right  angles  to  the  pipe,  the  height  at 


FIG.  81. 

which  the  water  stands  in  any  tube  represents  the  kinetic  pressure 
head  at  this  point.  Assuming  that  the  pipe  is  straight  and  of 
uniform  cross  section,  the  velocity  head  is  constant  throughout, 
and  therefore  as  the  frictional  head  increases  the  pressure  head 
decreases.  The  head  lost  in  friction  between  any  two  points 
m  and  n  (Fig.  81)  as  given  by  Eq.  (48),  Art.  17,  is 


and  is  therefore  proportional  to  the  distance  I  between  these 
points.  Consequently,  the  drop  in  the  piezometer  column  be- 
tween any  two  points  is  proportional  to  their  distance  apart,  and 


HYDROKINETICS 


83 


therefore  the  tops  of  these  columns  must  lie  in  a  straight  line. 
This  line  is  called  the  hydraulic  gradient,  or  virtual  slope  of  the 
pipe.  Evidently  the  vertical  ordinate  between  any  point  in  the 
pipe  and  the  hydraulic  gradient  measures  the  kinetic  pressure 
head  at  the  point  in  question. 

Slope  of  Hydraulic  Gradient. — When  a  pipe  is  not  straight, 
successive  points  on  the  hydraulic  gradient  may  be  determined  by 
computing  the  loss  of  head  between  these  points  from  the  relation 


taking  as  successive  values  of  I  the  length  of  pipe  between  the 
points  considered. 

In  water  mains  the  vertical  curvature  of  the  pipe  line  is  gen- 
erally small,  and  its  effect  on  the  hydraulic  gradient  is  usually 


A- 


FIG.  82. 

neglected.  When,  however,  a  valve  or  other  obstruction  occurs 
in  a  pipe  there  is  a  sudden  drop  in  the  hydraulic  gradient  at  the 
obstruction,  due  to  the  loss  of  head  caused  by  it. 

It  should  be  noted  that  the  upper  end  of  the  hydraulic  gradient 
lies  below  the  water  level  in  the  reservoir  a  distance  equal  to  the 
head  lost  at  entrance  plus  the  velocity  head.  The  slope  of  the 
hydraulic  gradient  is  usually  defined,  however,  as 

,       ,       ..  static  head 

Slope  of  hydraulic  gradient  =  , TT — r — : — > 

length  of  pipe 

which  is  equivalent  to  neglecting  the  velocity  head  and  head  lost 
at  entrance,  thereby  making  the  assumed  hydraulic  gradient 
slightly  steeper  than  it  actually  is. 

Peaks  above  Hydraulic  Gradient. — When  part  of  the  pipe 
line  rises  above  the  hydraulic  gradient  (Fig.  82),  the  pressure  in 
this  portion  must  be  less  than  atmospheric  since  the  pressure 
head  h'  becomes  negative.  If  the  pipe  is  air-tight  and  filled  be- 


84  ELEMENTS  OF  HYDRAULICS 

fore  the  flow  is  started  this  will  not  affect  the  discharge.  If  the 
pipe  is  not  air-tight,  air  will  collect  at  the  summit  above  the 
hydraulic  gradient,  changing  the  slope  of  the  latter  from  AB  to 
AC  as  indicated  in  Fig.  82,  thereby  reducing  the  head  to 
h"  with  a  corresponding  diminution  of  the  flow.  Before  laying 
a  long  pipe  line  the  hydraulic  gradient  should  therefore  be  plotted 
on  the  profile  to  make  sure  there  are  no  summits  projecting  above 
the  gradient.  In  case  such  summits  are  unavoidable,  provision 
should  be  made  for  exhausting  the  air  which  may  collect  at  these 
points,  so  as  to  maintain  full  flow. 

19.  HYDRAULIC  RADIUS 

Definition.  —  That  part  of  the  boundary  of  the  cross  section  of  a 
channel  or  pipe  which  is  in  contact  with  the  water  in  it  is  called 
the  wetted  perimeter,  and  the  area  of  the  cross  section  of  the  stream 
divided  by  the  wetted  perimeter  is  called  the  hydraulic  radius,  or 
hydraulic  mean  depth.  lii  what  follows  the  hydraulic  radius  will 
be  denoted  by  r,  defined  as 

Area  of  flow 

Hydraulic  radius,  r  =  =^r—         —  r—    — 
Wetted  perimeter 

Some  writers  apply  the  term  hydraulic  radius  only  to  circular 
pipes,  and  use  the  term  hydraulic  mean  depth  for  flow  in  channels. 
For  a  channel  of  rectangular  cross  section  having  a  breadth  6 
and  depth  of  water  h,  the  hydraulic  radius  is 

bh 
~  b  +  2h 

In  a  circular  pipe  of  diameter  d,  running  full,  the  hydraulic 
radius  is 


r 
For  the  same  pipe  running  half  full, 


ird         4 


8          d 
r-=^d==-*' 
2 

and  is  therefore  the  same  as  when  the  pipe  is  full. 


HYDROKINETICS  85 

Other  examples  of  the  hydraulic  radius  are  shown  in  Figs. 
98-104. 

Chezy's  Formula.  —  The  formula  proposed  by  Chezy  for  the 
velocity  of  flow  in  a  long  pipe  is 


v  =  C  Vrs^  (56) 

where  s  denotes  the  slope  of  the  hydraulic  gradient,  defined  in  the 
preceding  article;  r  is  the  hydraulic  radius,  defined  above;  and 
C  is  an  empirical  constant  which  depends  on  the  velocity  of  flow, 
diameter  of  pipe,  and  roughness  of  its  lining. 

For  a  circular  pipe  flowing  full  Chezy  's  formula  is  identical 
with  the  formula  for  friction  loss  in  a  pipe,  given  by  Eq.  (48), 
Art.  17,  namely, 


To  show  this  identity,  substitute  in  Chezy 's  formula  the  values 


d       .          h 
r  =  -7  and  s  =  T- 


Then  it  becomes 


whence,  by  squaring  and  solving  for  h,  it  takes  the  form 
4  il\  Sgil 


Consequently,  if  the  constant  term  ™  is  denoted  by  /,  that  is, 


Chezy  's  formula  assumes  the  standard  form, 


The  most  important  application  of  Chezy's  formula  is  to  flow 
in  open  channels,  as  explained  in  Art.  24. 

It  has  been  found  by  experiment  that  the  coefficient  C  in 
Chezy  's  formula  is  not  strictly  constant  for  any  particular  chan- 
nel and  dependent  only  on  the  roughness  of  the  channel  lining, 
but  that  it  also  varies  with  the  slope  and  the  hydraulic  radius. 


86 


ELEMENTS  OF  HYDRAULICS 


This  variation  may  be  taken  into  account  by  writing  the  formula 
in  the  exponential  form 

v  =  Crmsn 

where  the  exponents  m  and  n  are  also  empirical  constants.  Ex- 
perimental data  for  this  formula  are  not  yet  sufficient,  however,  to 
make  its  use  practical. 

20.  DIVIDED  FLOW 

Compound  Pipes. — In  water  works  calculations  the  problem 
often  arises  of  determining  the  flow  through  a  compound  system 
of  branching  mains. 

To  illustrate  the  method  of  finding  the  discharge  through  the 
various  branches,  consider  first  the  simple  case  of  a  main  tapped 


f 


M 


B 


N 


Plan 


FIG.  83. 


by  a  branch  pipe  which  later  returns  to  the  main,  as  indicated  in 
Fig.  83.  The  solution  in  this  case  is  based  on  the  fundamental 
relation  deduced  in  Art.  17,  namely, 


where  h  denotes  the  static  head,  and  hi  the  head  lost  in  friction. 
Using  the  notation  indicated  on  the  figure  and  considering  the 


HYDROKINETICS 


87 


d2g 


two  branches  separately,  we  obtain  the  following  equations :- 
For  line  ABMCD, 


For  line  ABNCD, 


By  subtraction  of  these  two  equations  we  have 


which  shows  that  the  frictional  head  lost  in  the  branch  BMC  is 
equal  to  that  lost  in  BNC. 

a 

nf 


FIG.  84. 

Since  the  total  discharge  through  the  branches  is  the  same  as 
that  through  the  main  before  dividing  and  after  uniting,  we  also 
have  the  two  relations 

i  ft\f\\ 

QiVi   —   G2V2  ~T~  ttsVs   —   <Z4i>4.  WW 

By  assuming  an  average  value  for  the  frictional  coefficient  /, 
the  four  Eq.  (57),  (58),  (59)  and  (60)  may  then  be  solved  for 
the  four  unknowns  Vi,  v2,  v3j  v±.  Having  found  approximate 
values  of  the  velocities,  corresponding  values  of  /  may  be  sub- 
stituted in  these  equations  and  the  solution  repeated,  thus  giving 
more  accurate  values  of  the  velocities. 

Having  found  the  velocities,  the  discharge  through  the  various 
pipes  may  be  obtained  from  the  relations 
r\        f\ 

Icgi    ==    t^/4    — 


88  ELEMENTS  OF  HYDRAULICS 

The  solution  for  more  complicated  cases  is  identical  with  the 
above,  except  that  more  equations  are  involved. 

Branching  Pipes.  —  Another  simple  case  of  divided  flow  which 
is  often  met  is  that  in  which  a  pipe  AB  of  diameter  d  divides  at 
some  point  B  into  two  other  pipes,  BC  and  BD,  of  diameters 
di  and  d2  respectively,  which  discharge  into  reservoirs  or  into  the 
air  (Fig.  84).  If  any  outlet,  as  C,  is  higher  than  the  junction  B, 
then  in  order  for  flow  to  take  place  in  the  direction  BC,  the  hy- 
draulic gradient  must  slope  in  this  direction;  that  is  to  say  there 
must  be  a  drop  in  pressure  between  the  junction  B  and  the  level 
of  the  outlet  reservoir  C,  or,  in  the  notation  of  the  figure,  the  con- 
dition for  flow  in  the  direction  BC  is  hi  >  h. 

Assuming  this  to  be  the  case,  the  solution  is  obtained  from  the 
same  fundamental  relation  as  above,  namely, 


Using  the  notation  indicated  on  the  figure  for  length,  diameter 
and  velocity  in  the  various  pipes  and  considering  one  line  at  a 
time,  we  thus  obtain  the  following  equations: 


ForlineABC,      ^  =        +  /  +/'  (*> 

Forline^Z),      *,- 

Also,  from  the  condition  that  the  discharge  through  the  main 
pipe  must  equal  the  sum  of  the  discharges  through  the  branches, 
denoting  the  cross-sectional  areas  by  a,  ai,  «2  respectively,  we 
have 

av  =  aiVi  +  a2v2.  (49) 

By  assuming  an  average  value  for  the  frictional  coefficient  /, 
these  three  equations  may  then  be  solved  for  the  three  unknowns 
v,  Vi  and  v%.  Having  thus  found  approximate  values  of  the  veloc- 
ities, the  exact  value  of  /  corresponding  to  each  velocity  may  be 
substituted  in  the  above  equations  and  the  solution  repeated, 
giving  more  accurate  values  of  the  velocities.  Having  found  the 
velocities,  the  discharge  from  each  pipe  is  obtained  at  once  from 
the  relations 

Q  =  av;    Qi  =  a^i;    Q2  =  a2v2. 


HYDROKINETICS  89 

The  method  of  solution  is  the  same  for  any  number  of  branches, 
there  being  as  many  equations  in  any  given  case  as  there  are  un- 
known velocities  to  be  determined. 

Other  simple  cases  of  divided  flow  are  illustrated  in  the  numer- 
ical examples  at  the  end  of  the  chapter. 

21.  FIRE  STREAMS 

Freeman's  Experiments.  —  Extensive  and  accurate  experi- 
ments on  discharge  through  fire  hose  and  nozzles  were  made  by 
John  R.  Freeman  at  Lawrence,  Mass.,  in  1888  and  1890.  1 

From  these  experiments  it  was  found  that  the  smooth  cone 
nozzle  with  simple  play  pipe  is  the  most  efficient  for  fire  streams, 
the  coefficient  of  discharge  being  nearly  constant  for  the  various 
types  tried  and  having  an  average  value  of  0.974  for  smooth  cone 
nozzles  and  0.74  for  square  ring  nozzles. 

The  friction  losses  for  fire  hose  were  found  to  depend  chiefly  on 
the  nature  of  the  interior  surface  of  the  hose,  the  loss  in  unlined 
linen  hose  being  about  2-1/2  times  as  great  as  in  smooth  rubber 
hose  of  the  same  diameter.  The  friction  loss  was  also  found  to 
vary  approximately  as  the  square  of  the  velocity  of  flow. 

For  fire  hose  laid  in  ordinary  smooth  curves  but  not  cramped  or 
kinked,  the  friction  loss  was  found  to  be  about  6  per  cent,  greater 
than  in  perfectly  straight  hose. 

Formulas  for  Discharge.  —  The  following  formulas  for  discharge 
were  deduced  by  Freeman  from  these  experiments. 
Notation: 

Q  =  discharge  in  cubic  feet  per  second, 

G  =  discharge  in  gallons  per  minute  =  448.83Q, 
h  =  piezometer  reading  at  base  of  nozzle  in  feet  of  water, 
p  =  pressure  at  base  of  nozzle  in  pounds  per  square  inch  = 
0.434/i, 

K  =  coefficient  of  discharge, 

Cc  =  coefficient  of  contraction, 
d  =  diameter  of  nozzle  orifice  in  inches, 

D  =  diameter  of  channel,  where  pressure  is  measured,  in  inches, 

H  =  total  hydrostatic  head  in  feet  =  - 


1  Trans.  Am.  Soc.  C.  E.  Vol.  21,  pp.  303-482;  Vol.  24,  pp.  492-527. 


90 
Then 


ELEMENTS  OF  HYDRAULICS 


Q  =  o.04374Kd2 


and 


o.o6645Kd 


(63) 


G= 


i  -K' 


d\4 


=  29.83 


i  -K2  ^ 


(64) 


Height  of  Effective  Fire  Stream. — It  was  also  found  that  the 
height,  y,  of  extreme  drops  in  still  air  from  nozzles  ranging  in  size 
from  3/4  in.  to  1-3/8  in.  in  diameter  was  given  by  the  formula 


H2 


y  =  H  -  0.00135^' 


(65) 


The  height  of  a  first  class  fire  stream  will  then  be  a  certain  frac- 
tion of  y  as  indicated  in  the  following  table: 


When  y  = 

50ft. 

75ft. 

100  ft. 

125  ft. 

150  ft. 

Height  of  first  class  fire  stream  = 

0.82  y 

0.79  y 

0.73  y 

0.67  y 

0.63  y 

Table  11  is  abridged  from  a  similar  table  computed  by  Free- 
man from  these  and  other  formulas,  not  here  given,  and  will  be 
found  convenient  to  use  in  solving  fire-stream  problems. 


22.  EXPERIMENTS  ON  THE  FLOW  OF  WATER 

Verification  of  Theory  by  Experiment. — The  subject  of  hydrau- 
lics as  presented  in  an  elementary  text  book  is  necessarily  limited 
to  simple  demonstrations  of  the  fundamental  principles.  It 
should  not  be  inferred  from  this  that  the  subject  is  largely  experi- 
mental and  not  susceptible  of  mathematical  analysis.  As  a  mat- 
ter of  fact,  hydrodynamics  is  one  of  the  most  difficult  branches  of 
applied  mathematics,  and  its  development  has  absorbed  the 


HYDROKINETICS 


91 


best  efforts  of  such  eminent  mathematical  physicists  as  Poinsot, 
Kirch  hoff,  Helmholz,  Maxwell,  Kelvin,  Stokes  and  Lamb.  Natur- 
ally the  results  are  too  technical  to  be  generally  appreciated,  but 
afford  a  rich  field  for  study  to  those  with  sufficient  mathematical 
preparation. 


Sudden  contraction. 


Sudden  enlargement. 


FIG.  85. 


Some  of  the  results  concerning  the  flow  of  liquids  derived  by 
mathematical  analysis  have  been  verified  experimentally  by  the 
English  engineers,  Professor  H.  S.  Hele-Shaw  and  Professor 
Osborne  Reynolds.  The  chief  importance  of  these  experiments  is 
that  they  serve  to  visualize  difficult  theoretical  results. 


Contraction.  Enlargement. 

FIG.  86. 

Method  of  Conducting  Experiments. — In  Art.  8  a  stream  line 
was  defined  as  the  path  followed  by  a  particle  of  liquid  in  its 
motion.  A  set  of  stream  lines  distributed  through  a  flowing 
liquid  therefore  completely  determines  the  nature  of  the  flow. 
To  make  such  stream  lines  visible,  so  as  to  make  it  possible  to 
actually  trace  the  motion  of  the  particles  of  a  clear  fluid,  both 
experimenters  named  above  allowed  small  bubbles  of  air  to  enter 


92 


ELEMENTS  OF  HYDRAULICS 


a  flowing  stream.  These  bubbles  do  not  make  the  motion  directly 
visible  to  the  eye,  but  by  making  the  pipe  or  channel  of  glass  and 
projecting  a  portion  of  it  on  a  screen  by  means  of  a  lantern,  its 
image  on  the  screen  as  viewed  in  this  transmitted  light  clearly 
shows  certain  characteristic  features. 

Effect  of  Sudden  Contraction  or  Enlargement. — Figs.  85  and  86, 
reproduced  by  permission  of  Professor  Hele-Shaw,  show  the 
effect  of  a  sudden  contraction  or  enlargement  of  the  channel  sec- 
tion. It  is  noteworthy  that  the  disturbance  or  eddying  is  much 


FIG.  87. 

greater  for  a  sudden  enlargement  than  for  a  sudden  contraction. 
This  is  due  to  the  inertia  of  the  fluid  which  prevents  it  from  imme- 
diately filling  the  channel  after  passing  through  the  orifice.  This 
also  confirms  what  has  already  been  observed  in  practice, 
namely,  that  the  loss  of  energy  due  to  a  sudden  enlargement 
in  a  pipe  is  much  greater  than  that  due  to  a  corresponding 
contraction. 

Disturbance  Produced  by  Obstacle  in  Current. — If  the  channel 
is  of  considerable  extent  and  a  small  obstacle  is  placed  in  it,  the 
stream  lines  curve  around  the  obstacle,  leaving  a  small  space 
behind  it,  as  shown  in  Fig.  87.  If  the  object  is  a  square  block  or 
flat  plate  this  effect  is  greatly  magnified,  as  shown  in  Fig.  88. 
The  water  is  prevented  from  closing  at  once  behind  the  obstacle 
by  reason  of  its  inertia.  This  indicates  why  the  design  of  the 


HYDROKINETICS 


93 


stern  of  a  ship  is  so  much  more  important  than  that  of  the  bow, 
since  if  there  are  eddys  in  the  wake  of  "a  ship,  the  pressure  of  the 
water  at  the  stern  is  decreased,  thereby  increasing  by  just  this 
much  the  effective  resistance  to  motion  at  the  bow. 

Stream -line  Motion  in  Thin  Film. — In  these  experiments  it  was 
also  observed  that  there  was  always  a  clear  film  of  liquid,  or 
border  line,  on  the  sides  of  the  channel  and  around  the  obstacle. 
This  observed  fact  was  accounted  for  on  the  ground  that  by 
reason  of  the  friction  between  a  viscous  liquid  and  the  sides  of  the 


FIG.  88. 

channel  or  obstacle,  the  thin  film  of  liquid  affected  was  not  mov- 
ing with  turbulent  motion  but  with  true  stream  line  motion,  as 
in  an  ideal  fluid.  To  isolate  this  film  so  as  to  observe  its  motion, 
water  was  allowed  to  flow  between  two  plates  of  glass  in  a  sheet 
so  thin  that  its  depth  corresponded  to  the  thickness  of  the  border 
line  previously  observed.  When  this  was  done  it  immediately 
became  apparent  that  the  flow  was  no  longer  turbulent  but  a 
steady  stream-line  motion.  The  flow  of  a  viscous  fluid  like  gly- 
cerine in  a  thin  film  thus  not  only  eliminates  turbulent  flow,  but 
also  to  a  certain  extent  the  inertia  effects,  thereby  resulting  in 
true  stream-line  flow. 

Cylinder  and  Flat  Plate. — To  make  the  stream  lines  visible, 
colored  liquid  was  injected  through  a  series  of  small  openings,  the 
result  being  to  produce  an  equal  number  of  colored  bands  or 
stream  lines  in  the  liquid.  Fig.  89  shows  these  stream-lines  for 


94 


ELEMENTS  OF  HYDRAULICS 


FIG.  89. 


FIG.  90. 


Theory 


Experiment 


HYDROKINETICS 


95 


a  cylinder,  and  Fig.  90  for  a  flat  plate  placed  directly  across  the 
current,  while  Fig.  91  shows  a  comparison  of  theory  and  experi- 
ment for  a  flat  plate  inclined  to  the  current. 

Velocity  and  Pressure. — The  variation  in  thickness  of  the 
bands  is  due  to  the  difference  in  velocity  in  various  parts  of  the 
channel,  the  bands  of  course  being  thinnest  where  the  velocity  is 
greatest.  Since  a  decrease  in  velocity  is  accompanied  by  a  cer- 
tain increase  in  pressure,  the  wide  bands  before  and  behind  the 
obstacle  indicate  a  region  of  higher  pressure.  This  accounts  for 
the  standing  bow  and  stern  waves  of  a  ship,  whereas  the  narrow- 
ing of  the  bands  at  the  sides  indicates  an  increase  of  velocity  and 
reduction  of  pressure,  and  accounts  for  the  depression  of  the  water 
level  at  this  part  of  a  ship. 


FIG.  92. 

In  the  case  of  a  sudden  contraction  or  enlargement  of  the  chan- 
nel section,  the  true  stream-line  nature  of  the  flow  was  clearly 
apparent,  as  shown  in  Fig.  92,  the  stream  lines  following  closely 
the  form  derived  by  mathematical  analysis  for  a  perfect  fluid. 


23.  MODERN  SIPHONS 

Principle  of  Operation. — In  its  simplest  form,  a  siphon  is  merely 
an  inverted  U-shaped  tube,  with  one  leg  longer  than  the  other, 
which  is  used  for  emptying  tanks  from  the  top  when  no  outlet 
is  available  below  the  surface.  In  use,  the  tube  is  filled  with 
liquid  and  the  ends  corked,  or  otherwise  closed.  The  short  end 
of  the  tube  is  then  placed  in  the  reservoir  to  be  lowered,  so  that 
the  level  of  the  end  outside  is  lower  than  the  surface  of  the 
reservoir  (Fig.  93).  When  the  ends  of  the  tube  are  opened,  the 
liquid  in  the  reservoir  begins  to  flow  through  the  tube  with  a  head, 


96 


ELEMENTS  OF  HYDRAULICS 


T 


h,  equal  to  the  difference  in  level  between  the  surface  of  the 
reservoir  and  the  lower,  or  outer,  end  of  the  tube.  If  the  inner 
end  of  the  siphon  is  placed  close  to  the  bottom  of  the  reservoir 

it  can  be  practically 
emptied  in  this  manner. 
For  emptying  small  tanks 
a  siphon  can  conveniently 
be  made  of  a  piece  of  ordi- 
nary tubing  or  hose. 

Siphon  Spillway. — Re- 
cent commercial  applica- 
tions of  the  siphon  on  a 
large  scale  are  the  siphon 
spillways  and  siphon  lock 
93  on  the  New  York  State 

barge  canal.  At  three  lo- 
calities on  the  Champlain  division  of  the  Barge  canal  the  siphon 
principle  is  being  used  for  the  first  time  to  create  a  spillway  of  any 
considerable  size.  At  the  place  shown  in  Fig.  94,  it  was  necessary 


_L 


PIG.  94. — Siphon   Spillway,  Champlain  division,  New  York  State  barge 

canal. 

to  provide  for  a  maximum  outflow  of  about  700  cu.  ft.  per  second 
and  to  limit  the  fluctuation  of  water  surface  to  about  1  ft.  The 
ordinary  waste  weir  of  a  capacity  sufficient  to  take  care  of  this 


HYDROKINETICS 


97 


98 


ELEMENTS  OF  HYDRAULICS 


HYDROKINETICS 


99 


flow,  with  a  depth  of  only  1  ft.  of  water  on  the  crest,  would  re- 
quire a  spillway  200  ft.  long.     The  siphon  spillway  measures 


only  57  ft.  between  abutments  and  accomplishes  the  same  results. 
This  particular  structure  consists  of  four  siphons,  each  having  a 


100  ELEMENTS  OF  HYDRAULICS 

cross-sectional  area  of  7-3/4  sq.  ft.  and  working  under  a  10-1/2-ft. 
head.  There  is  also  a  20-ft.  drift  gap  to  carry  off  floating  debris. 
The  main  features  of  construction  are  shown  in  Fig.  95.  The 
siphon  spillway  was  designed  and  patented  by  Mr.  George  F. 
Stickney,  one  of  the  Barge  canal  engineers. 

Siphon  Lock. — The  siphon  lock  on  the  New  York  State  Barge 
Canal  is  located  in  the  city  of  Oswego,  and  is  the  only  lock  of  this 
type  in  this  country  and  the  largest  ever  built  on  this  principle. 
It  consists  of  two  siphons,  as  shown  in  Fig.  96,  the  crown  of  each 
being  connected  by  a  4-in.  pipe  to  an  air  tank  in  which  a  partial 
vacuum  is  maintained.  To  start  the  flow,  the  air  valve  is  opened, 
the  vacuum  in  the  tank  drawing  the  air  from  the  siphon  and  there- 
by starting  the  flow.  When  the  siphon  is  discharging  fully,  its 
draft  is  such  that  the  air  is  sucked  out  of  the  tank,  thus  restoring 
the  partial  vacuum.  To  stop  the  flow,  outer  air  is  admitted  to 
the  crest  of  the  siphon  by  another  valve,  thereby  breaking  the 
flow,  as  indicated  in  Fig.  97.  The  operating  power  is  thus  self 
renewing,  and;  except  for  air  leakage,  lockages  can  be  conducted 
by  merely  manipulating  the  4-in.  air  valves.  However,  to  avoid 
the  necessity  of  refilling  the  tank  when  traffic  is  infrequent,  it  is 
customary  to  close  the  20-in.  outlet  valve,  thus  holding  the 
water  in  the  tank.  Using  both  siphons,  the  lock  chamber  can 
be  filled  in  4-1/2  to  5  min.,  and  emptied  in  5-1/2  to  6  min. 

24.  FLOW  IN  OPEN  CHANNELS 

Open  and  Closed  Conduits. — Conduits  for  conveying  water 
are  usually  classified  as  open  and  closed.  By  a  closed  conduit  is. 
meant  one  flowing  full  under  pressure,  as  in  the  case  of  ordinary 
pipe  flow  discussed  in  Art.  15.  Water  mains,  penstocks,  draft 
tubes  and  fire  hose  are  all  examples  of  closed  conduits. 

Open  channels,  or  conduits,  are  those  in  which  the  upper  sur- 
face of  the  liquid  is  exposed  to  atmospheric  pressure  only,  the 
pressure  at  any  point  in  the  stream  depending  merely  on  the 
depth  of  this  point  below  the  free  surface.  Rivers,  canals, 
flumes,  aqueducts  and  sewers  are  ordinarily  open  channels.  A 
river  or  canal,  however,  may  temporarily  become  a  closed  chan- 
nel when  covered  with  ice,  and  an  aqueduct  or  sewer  may  also 
become  a  closed  channel  if  flowing  full  under  pressure. 

Steady  Uniform  Flow. — The  fundamental  laws  applying  to 
flow  in  open  and  closed  channels  are  probably  identical,  and  in 


HYDROKINETICS 


101 


the  case  of  steady,  uniform  flow  the  same  formulas  apply  to  both. 
For  steady  flow  in  an  open  channel  the  quantity  of  water  passing 
any  transverse  section  of  the  stream  is  constant,  and  for  uniform 
flow  the  mean  velocity  is  also  constant.  Under  these  conditions 
the  cross-sectional  area  of  the  stream  is  constant  throughout  its 
length,  and  the  hydraulic  gradient  is  the  slope  of  the  surface  of  the 
stream.  The  formula  for  velocity  of  flow  is  then  the  one  given  in 
Art.  19  under  the  name  of  Chezy's  formula,  namely 

v  =  C  Vrs.  (66) 

Kutter's  Formula.  —  Numerous  experiments  have  been  made  to 
determine  the  value  of  the  coefficient  C  for  open  channels.  In 
1869,  E.  Ganguillet  and  W.  R.  Kutter,  two  Swiss  engineers,  made 
a  careful  determination  of  this  constant,  the  result  being  expressed 
in  the  following  form  : 


v  = 


0.00281         1.811 
41-65  +  -  + 


i  +    41.65  + 


Vrs 


(67) 


in  which 

s  =  hydraulic  gradient,  or  slope  of  channel, 

area  of  flow 

r  =  hydraulic  radius  =  -      —  ,  --  ~.  —     —  > 

wetted  perimeter 

n  =  coefficient  of  roughness. 

The  coefficient  of  roughness,  n,  depends  on  the  nature  of  the 
channel  lining.  Approximate  values  of  n  for  various  surfaces  are 
given  in  the  following  table: 


Nature  of  Channel  Lining 


Planed  timber  carefully  joined,  glazed  or  enameled  surfaces 

Smooth  clean  cement 

Cement  mortar,  one-third  sand 

Unplaned  timber  or  good  new  brickwork 

Smooth  stonework,  vitrified  sewer  pipe  and  ordinary  brickwork  .  . 

Rough  ashlar  and  good  rubble  masonry 

Firm  gravel 

Ordinary  earth 

Earth  with  stones,  weeds,  etc 

Earth  or  gravel  in  bad  condition 


0.009 
0.010 
0.011 
0.012 
0.013 
0.017 
0.020 
0.025 
0.030 
0.035 


Limitations  to  Kutter 's  Formula. — Kutter's  formula,  Eq.  (67), 
is  widely  used  and  is  reliable  when  applied  to  steady,  uniform 


102  ELEMENTS  OF  HYDRAULICS 

flow  under  normal  conditions.  From  a  study  of  the  data  on 
which  this  formula  is  based,  its  use  has  been  found  to  be  subject 
to  the  following  limitations: 

It  is  not  reliable  for  hydraulic  radii  greater  than  10  ft.,  or  ve- 
locities greater  than  10  ft.  per  sec.,  or  slopes  flatter  than  1  in 
10,000.  Within  these  limits  a  variation  of  about  5  per  cent,  may 
be  expected  between  actual  results  and  those  computed  from  the 
formula. 

Table  15  gives  numerical  values  of  the  coefficient  C  calculated 
from  Formula  (67). 

Bazin's  Formula.  —  In  1897,  H.  Bazin  also  made  a  careful 
determination  of  the  coefficient  C  from  all  the  experimental  data 
then  available,  as  the  result  of  which  he  proposed  the  following 
formula: 


Vr 

where    r  =  hydraulic  radius, 

m  =  coefficient  of  roughness. 

Bazin'  s  formula  has  the  advantage  of  being  simpler  than 
Kutter's,  and  is  independent  of  the  slope  s.  Values  of  the  coeffi- 
cient of  roughness,  m,  for  use  with  this  formula  are  given  in  the 
following  table: 


Nature  of  Channel  Lining 


Planed  timber  or  smooth  cement 

Unplaned  timber,  well-laid  brick  or  concrete 

Ashlar,  good  rubble  masonry  or  poor  brickwork 

Earth  in  good  condition 

Earth  in  ordinary  condition 

Earth  in  bad  condition . . 


0.06 
0.16 
0.46 
0.85 
1.30 
1.75 


Table  14  gives  numerical  values  of  the  coefficient  C  calculated 
from  formula  (68). 

Kutter's  Simplified  Formula. — A  simplified  form  of  Kutter's 
formula  which  is  also  widely  used  is  the  following: 


f  100  Vr  1    , 
v  =    -  --  -=.    V 
Lb  +  Vr  J 


rs, 


where  b  is  a  coefficient  of  roughness  which  varies  from  0.12  to  2.44. 
For  ordinary  sewer  work  the  value  of  this  coefficient  may  be 
assumed  as  b  =  0.35. 


HYDROKINETICS  103 

26.  CHANNEL  CROSS  SECTION 

Condition  for  Maximum  Discharge. — From  the  Chezy  formula 
for  flow  in  open  channels,  namely, 

Q  =  Av  =  ACVrs, 

it  is  evident  that  for  a  given  stream  section  A  and  given  slope  s, 
the  maximum  discharge  will  be  obtained  for  that  form  of  cross 
section  for  which  the  hydraulic  radius  r  is  a  maximum.  Since 

area  of  flow 


wetted  perimeter 

this  condition  means  that  for  constant  area  the  radius  r,  and 
therefore  the  discharge,  is  a  maximum  when  the  wetted  perimeter 
is  a  minimum.  The  reason  for  this  is  simply  that  by  making  the 
area  of  contact  between  channel  lining  and  water  as  small  as 
possible,  the  frictional  resistance  is  reduced  to  a  minimum,  thus 
giving  the  maximum  discharge. 

Maximum  Hydraulic  Efficiency. — In  consequence  of  this,  it 
follows  that  the  maximum  hydraulic  efficiency  is  obtained  from 
a  semicircular  cross  section,  since  for  a  given  area  its  wetted 
perimeter  is  less  than  for  any  other  form  (Fig.  98) .  For  rectangu- 
lar sections  the  half  square  has  the  least  perimeter  for  a  given 
area,  and  consequently  is  most  efficient  (Fig.  99).  Similarly, 
for  a  trapezoidal  section  the  half  hexigon  is  the  most  efficient 
(Fig.  100).  In  each  case  the  hydraulic  radius  is  half  the  water 
depth,  as  proved  below. 

In  the  case  of  unlined  open  channels  it  is  necessary  to  use  the 
trapezoidal  section,  the  slope  of  the  sides  being  determined  by 
the  nature  of  the  soil  forming  the  sides.  This  angle  having  been 
determined,  the  best  proportions  for  the  section  are  obtained  by 
making  the  sides  and  bottom  of  the  channel  tangent  to  a  semi- 
circle drawn  with  center  in  the  water  surface  (Fig.  101). 

Regular  Circumscribed  Polygon. — Any  section  which  forms 
half  of  a  regular  polygon  of  an  even  number  of  sides,  and  has  each 
of  its  faces  tangent  to  a  semi-circle  having  its  center  in  the  water 
surface,  will  have  its  hydraulic  radius  equal  to  half  the  radius  of 
this  inscribed  circle  (Figs.  98-103).  To  prove  this,  draw  radii 
from  the  center  of  the  inscribed  circle  to  each  angle  of  the  polygon. 
Then  since  the  area  of  each  of  the  triangles  so  formed  is  equal  to 


104 


ELEMENTS  OF  HYDRAULICS 


Semicircle 
FIG.  98. 


Half  Square 
FIG.  99. 


4=3#tan  30° 
w.p.  =  6  R  tan   30° 

r_8Jg2  tan  30°  _R 
QR    tan  30°       2 


4/2  tan  30 


P=  Wetted  Perimeter 
A=!/2   PR 

=y±?R  =  R 


'2/2  tan  30 ol 
Half  Hexagon 

FIG.  100/ 


Trapezoid 
FIG.  101. 


P=  Wetted  Perimeter 
A=  V2  PR 


Half  Octagon 
FIG.  102. 


Triangular  Section 
FIG.  103. 


HYDROKINETICS 


105 


II- 


a- 


(N 


a  go 


04 


04 


04 


.a  -g 

c3  ""cu 


04 


04 


a  SD 

.§  -3 
'1 .1 


I 

00 
O 


04 


| 

CO 


O 


O  M 


106 


ELEMENTS  OF  HYDRAULICS 


one-half  its  base  times  its  altitude,  and  since  the  altitude  in  each 
case  is  a  radius  of  the  inscribed  circle,  the  total  area  is 

D 

Area  =  ^  X  perimeter. 
Consequently  the  hydraulic  radius  r  is 

area  of  flow          *  X  perimeter      R 


wetted  perimeter          perimeter  2 

Properties  of  Circular  and  Oval  Sections. — For  circular  and 
oval  cross  sections,  the  maximum  velocity  and  maximum  dis- 


Angle  0  in    Radians 


Circular  Section 

FIG.  104. 


i     Standard  Oval  Section 
FIG.  105. 


charge  are  obtained  when  the  conduit  is  flowing  partly  full,  as 
apparent  from  the  table  on  page  105,  which  is  a  collection  of  the 
most  important  data  for  circular  and  oval  sections,1  as  shown  in 
Figs.  104  and  105. 

Theoretically,  the  maximum  discharge  for  a  circular  pipe  oc- 
curs when  the  pipe  is  filled  to  a  depth  of  0.949  D,  but  if  it  is 
attempted  to  maintain  flow  at  this  depth,  the  waves  formed  in 
the  pipe  strike  against  the  top,  filling  it  at  periodic  intervals 
and  thus  producing  impact  losses.  To  obtain  the  maximum 
discharge'  without  danger  of  impact,  the  actual  depth  of  flow 
should  not  exceed  5/6  D. 

26.  FLOW  IN  NATURAL  CHANNELS 

Stream  Gaging. — In  the  case  of  a  stream  flowing  in  a  natural 
channel  the  conditions  determining  the  flow  are  so  variable  that 

1  Weyrauch,  Hydrauliches  Rechnen,  S.  51. 


HYDROKINETICS 


107 


no  formula  for  computing  the  discharge  has  been  devised  that  can 
claim  to  give  results  even  approximately  correct.  To  obtain 
accurate  results,  direct  measurements  of  cross  sections  and  veloc- 
ities must  be  made  in  the  field. 

The  two  methods  of  direct  measurement  in  general  use  are  as 
follows :     Either 


FIG.  106. — Electric  current  meter. 

1.  The  construction  of  a  weir  across  the  stream,  and  the  calcula- 
tion of  the  discharge  from  a  weir  formula;  or,  if  this  is  not  feasible, 

2.  The  measurement  of  cross  sections  of  the  stream  by  means  of 
soundings  taken  at  intervals,  and  the  determination  of  average 
velocities  by  a  current  meter  or  floats. 

The  first  of  these  methods  is  explained  in  Art.  11. 


108 


ELEMENTS  OF  HYDRAULICS 


Current  Meter  Measurements. — The  current  meter,  one  type 
of  which  is  shown  in  Fig.  106,  consists  essentially  of  a  bucket 
wheel  with  a  heavy  weight  suspended  from  it  to  keep  its  axis 
horizontal,  and  a  vane  to  keep  it  directed  against  the  cur- 
rent, together  with  some  form  of  counter  to  indicate  the  speed 
at  which  the  wheel  revolves.  The  meter  is  first  rated  by  towing 
it  through  still  water  at  various  known  velocities  and  tabulating 
the  corresponding  wheel  speeds.  From  these  results  a  table,  or 
chart,  is  constructed  giving  the  velocity  of  the  current  corre- 
sponding to  any  given  speed  of  the  wheel  as  indicated  by  the 
counter.  This  method  of  calibration,  however,  is  more  or  less 
inaccurate,  as  apparent  from  Du  Buat's  paradox,  explained  in 
Art.  27. 

Float  Measurements. — When  floats  are  used  to  determine  the 
velocity,  a  uniform  stretch  of  the  stream  is  selected,  and  two  cross 
sections  chosen  at  a  known  distance  apart.  Floats  are  then  put 

Water  Surface 


t 


Average 


Velocity 


Bed  ofyStream 


into  the  stream  above  the  upper  section  and  their  times  of  transit 
from  one  section  to  the  other  observed  by  means  of  a  stop  watch. 
A  sub-surface  float  is  commonly  used,  so  arranged  that  it  can  be 
run  at  any  desired  depth,  its  position  being  located  by  means  of  a 
small  surface  float  attached  to  it. 

If  the  cross  section  of  the  stream  is  fairly  uniform,  rod  floats 
may  be  used.  These  consist  of  hollow  tubes,  so  weighted  as  to 
float  upright  and  extend  nearly  to  the  bottom.  The  velocity 
of  the  float  may  then  be  assumed  to  be  equal  to  the  mean  velocity 
of  the  vertical  strip  through  which  it  runs. 

.  Variation  of  Velocity  with  Depth. — The  results  of  such  measure- 
ments show,  in  general,  that  the  velocity  of  a  stream  is  greatest 
midway  between  the  banks  and  just  beneath  the  surface.  In 


HYDROKINETICS 


109 


particular,  the  velocities  at  different  depths  along  any  vertical 
are  found  to  vary  as  the  ordinates  to  a  parabola,  the  axis  of  the 
parabola  being  vertical  and  its  vertex  just  beneath  the  surface,  as 
indicated  in  Fig.  107.  From  this  relation  it  follows  that  if  a 
float  is  adjusted  to  run  at  about  0.6  of  the  depth  in  any  vertical 
strip,  it  will  move  with  approximately  the  average  velocity  of  all 
the  particles  in  the  vertical  strip  through  which  it  runs. 

Calculation  of  Discharge. — In  order  to  calculate  the  discharge 
it  is  necessary  to  measure  the  area  of  a  cross  section  as  well  as  the 
average  velocities  at  various  points  of  this  section.  The  total 
cross  section  is  therefore  subdivided  into  parts,  say  A\t  A2,  As, 


etc.  (Fig.  108),  the  area  of  each  being  determined  by  measuring 
the  ordinates  by  means  of  soundings.  The  average  velocity  for 
each  division  is  then  measured  by  one  of  the  methods  explained 
above,  and  finally  the  discharge  is  computed  from  the  relation 


Q  = 


+  A2v2  +  A3v3  + 


27.  THE  PITOT  TUBE 

Description  of  Instrument. — An  important  device  for  measur- 
ing the  velocity  of  flow  is  the  instrument  known  as  the  Pilot  tube. 
In  1732  Pitot  observed  that  if  a  small  vertical  tube,  open. at  both 
ends,  with  one  end  bent  at  a  right  angle,  was  dipped  in  a  current 
so  that  the  horizontal  arm  was  directed  against  the  current  as 
indicated  in  Fig.  109,  A,  the  liquid  rises  in  the  vertical  arm  to  a 
height  proportional  to  the  velocity  head.  The  height  of  the  col- 
umn sustained  in  this  way,  or  hydrostatic  head,  is  not  exactly 
equal  to  the  velocity  head  on  account  of  the  disturbance  created 
by  the  presence  of  the  tube.  No  matter  how  small  the  tube  may 
be,  its  dimensions  are  never  negligible,  and  its  presence  has  the 
effect  of  causing  the  filaments  of  liquid,  or  stream  lines,  to  curve 


110 


ELEMENTS  OF  HYDRAULICS 


around  it,  thereby  considerably  modifying  the  pressure.  Since 
the  column  of  liquid  in  the  tube  is  sustained  by  the  impact  of  the 
current,  this  arrangement  is  called  an  impact  tube. 

If  a  straight  vertical  tube  is  submerged,  or  a  bent  tube  having  its 
horizontal  arm  directed  transversely,  that  is,  perpendicularly,  to 
the  current,  the  presence  of  the  tube  causes  the  stream  lines  to 
turn  their  concavity  toward  the  orifice,  thereby  producing  a  suc- 
tion which  is  made  apparent  by  a  lowering  of  the  water  level  in 
this  tube,  as  shown  in  Fig.  109,  B.  In  the  case  of  the  bent  tube,  if 
the  horizontal  arm  is  directed  with  the  current,  as  shown  in  Fig. 


Impact 
Tube 


Suctionor 
Pressure  Tube 


Tube 


Direction  of  Flow 

FIG.  109. 

109,  C,  the  suction  effect  is  more  pronounced,  and  the  level  in  the 
tube  is  still  further  lowered.  When  the  horizontal  arm  of  a  bent 
tube  is  directed  with  the  current,  the  arrangement  is  called  a 
suction  or  trailing  tube. 

It  is  practically  impossible,  however,  to  obtain  satisfactory 
numerical  results  with  this  simple  type  of  Pitot  tube,  as  in  the 
case  of  flow  in  open  channels  the  free  surface  of  the  liquid  is  usu- 
ally disturbed  by  waves  and  ripples  and  other  variations  in  level, 
which  are  often  of  the  same  order  of  magnitude  as  the  quantities 
to  be  measured;  while  in  the  case  of  pipe  flow  under  pressure  there 
are  other  conditions  which  strongly  affect  the  result,  as  will  appear 
in  what  follows. 

Darcy's  Modification  of  Pitot's  Tube. — In  1850  Darcy  modified 
the  Pitot  tube  so  as  to  adapt  it  to  general  current  measurements. 
This  modification  consisted  in  combining  two  Pitot  tubes,  as 
shown  in  Fig.  110,  the  orifice  of  the  impact  tube  being  directed 


HYDROKINETICS 


111 


up  stream,  and  the  orifice  of  the  suction  tube  transverse  to  the 
current.  In  some  forms  of  this  apparatus,  the  suction  tube  is 
of  the  trailing  type,  that  is,  the  horizontal  arm  is  turned  directly 
down  stream. 

To  make  the  readings  more  accurate,  the  difference  in  eleva- 
tion of  the  water  in  the  two  tubes  is  magnified  by  means  of  a 
differential  gage,  as  shown  in  Fig.  110. 
Here  A  denotes  the  impact  tube  and  B 
the  suction  tube  (often  called  the  pressure 
tube),  connected  with  the  tubes  C  and 
D.  between  which  is  a  graduated  scale. 
After  placing  the  apparatus  in  the  stream 
to  be  gaged,  the  air  in  both  tubes  is  equally 
rarified  by  suction  at  F,  thereby  causing 
the  water  level  in  both  to  rise  propor- 
tional amounts.  The  valve  at  F  is  then 
closed,  also  the  valve  at  E,  and  the  appara- 
tus is  lifted  from  the  water  and  the  read- 
ing on  the  scale  taken. 

It  was  assumed  by  Pitot  and  Darcy  that 
the  difference  in  level  in  the  tubes  was  pro- 
portional to  the  velocity  head  ^-t  where  v 

denotes  the  velocity  of  the  current.     Call-         H~[ 
ing  hi  and  h%  the  difference  in  level,  that 
is,    the    elevation   or   depression     of   the  g — -^J 
water  in  the  impact  and  suction  tubes  re- 
spectively,  and  mi,    m%  the   constants  of 
proportionality,  we  have  therefore 


T 


D 


FIG.  110. 


If,  then,  h  denotes  the  difference  in  elevation  in  the  two  tubes 
(Fig.  110),  we  have 


The  velocity  v  is  therefore  given  in  terms  of  h  by  the  equation 

v  =  m  Vigh  (69) 

where 


m  = 


mi  +  w2 


112 


ELEMENTS  OF  HYDRAULICS 


The  coefficient  m  depends,  like  mi  and  m2,  on  the  form  and  dimen- 
sions of  the  apparatus,  and  when  properly  determined  is  a  con- 
stant for  each  instrument,  provided  that  the  conditions  under 
which  the  instrument  is  used  are  the  same  as  those  for  which  m 
was  determined. 
The  value  of  m  in  this  formula  has  been  found  to  vary  from  1 

to  as  low  as  0.7;  the  value  m  =  1  corresponding  to  h  =  ~-;    and 

y 

V2 

the  value  m  =  0.7  to  h  =  — 

9 

The  explanation  of  this  ap- 
parent discrepancy  is  given 
below  under  the  theory  of  the 
impact  tube. 

In  the  case  of  variable  ve- 
locity   of    flow    it   has   been 
shown    by  Rateau1  that  the 
Pitot,  or  Darcy,  tube  meas- 
_j_  ures   not  the   mean   velocity 

but  the  mean  of  the  squares 
of  the  velocities  at  the  point 
where  it  is  placed  during  the 
experiment.  To  obtain  the 
mean  velocity  it  is  necessary 

v2 

to  multiply  <=>-  by  a  coeffi- 
cient which  varies  according 
to  the  rate  of  change  of  the 
velocity  with  respect  to  -the 
time.  From  Rateau's  experi- 
ments this  coefficient  was 
found  to  vary  from  1.012  to 
1.37,  having  a  mean  value  of 
1.15.  This  corresponds  to  a  mean  value  for  m  of  0.93. 

Pitometer. — A  recent  modification  of  the  Pitot  tube  is  an 
instrument  called  the  Pitometer  (Fig.  111).  The  mouth-piece 
of  this  apparatus  consists  of  two  small  orifices  pointing  in  oppo- 
site directions  and  each  provided  with  a  cut- water,  as  shown  in  the 
figure.  When  in  use,  these  are  set  in  line  parallel  to  the  current, 


Cutwater 


FIG.  111. 


1  Annales  des  Mines,  Mars,  1898. 


HYDROKINETICS  113 

so  that  one  points  directly  against  the  current  and  the  other  with 
it.  The  differential  gage  used  with  this  instrument  consists  of  a 
U-tube,  one  arm  of  which  is  connected  with  one  mouthpiece  and 
the  other  arm  with  the  other  mouthpiece,  and  which  is  about 
half  filled  with  a  mixture  of  gasoline  and  carbon  tetrachloride, 
colored  dark  red.  The  formula  for  velocity  as  measured  by  this 
instrument  is  given  in  the  form 

v  =  k[2g(s  - 


where    k  =  empirical   constant  =  0.84  for  the    instrument   as 

manufactured  and  calibrated, 

s  =  specific  weight  of  the  tetrachloride  mixture  =  1.25, 
d  =  difference  in  elevation  in  feet  between  the  tops  of  the 
two  columns  of  tetrachloride. 

Inserting  these  numerical  values,  the  formula  reduces  to 

v  =  3.368  VS. 

It  is  claimed  that  velocities  as  low  as  1/2  ft.  per  second  can  be 
measured  with  this  instrument. 

Pitot  Recorders.  —  The  Pitot  meter  is  used  in  power  houses, 
pumping  stations  and  other  places  where  a  Venturi  tube  cannot 
be  installed,  and  is  invaluable  as  a  water-works  instrument  to 
determine  the  pipe  flow  in  any  pipe  of  the  system. 

A  recent  portable  type,  especially  adapted  to  this  purpose  is 
shown  in  Fig.  112.  This  instrument  is  34  in.  high,  weighs  75  lb., 
and  furnishes  a  chart  of  the  Bristol  type  which  are  averaged  with 
a  special  planimeter  furnished  with  the  instrument.  A  1-in. 
tap  in  the  water  main  is  required  for  inserting  the  Pitot  mouth- 
piece. 

It  is  claimed  that  these  instruments  have  a  range  from  1/2  ft. 
per  second  to  any  desired  maximum. 

Theory  of  the  Impact  Tube.  —  The  wide  variation  in  the  range 
of  coefficients  recommended  by  hydraulic  engineers  for  use  with 
the  Pitot  tube  can  be  accounted  for  only  on  the  ground  of  a  faulty 
understanding  of  the  hydraulic  principles  on  which  its  action  is 
based.  The  most  important  of  these  are  indicated  below,  with- 
out presuming  to  be  a  complete  exposition  of  its  action. 

It  will  be  shown  in  Art.  29  that  the  force  produced  by  the  im- 
pact of  a  jet  on  a  flat  plate  is  twice  as  great  as  that  due  to  the  hy- 


114 


ELEMENTS  OF  HYDRAULICS 


drostatic  head  causing  the  flow.     That  is  to  say,  if  the  theoretical 
velocity  of  a  jet  is  that  due  to  a  head  h,  where 


FIG.  112.  —  Pitot  recording  meter,  Simplex  Valve  and  Meter  Co. 

the  force  exerted  on  a  fixed  plate  by  the  impact  of  this  jet  is  equal 
to  that  due  to  a  hydrostatic  head  of  hf  =  2h,  in  which  case 


The  orifice  in  a  Pitot  tube  is  essentially  a  flat  plate  subjected  to 
the  impact  of  the  current.     Considering  only  the  impact  effect, 


HYDROKINETICS 


115 


therefore,  the  head  which  it  is  theoretically  possible  to  attain  in  a 
Pitot  tube  is 

h'  =  v-, 

9 
which  corresponds  to  a  value  of  m  of  0.7  in  the  formula 

v  —  m 


D' 


There  are  other  considerations,  however,  which  often  modify 
this  result  considerably.  The  effect  of  immersing  a  circular 
plate  in  a  uniform  parallel  current  has  been  fully  analyzed  theo- 
retically and  the  results  confirmed  experimentally.  The  re- 
sults of  such  an  analysis  made  by  Professor  Prasil,  as  pre- 
sented in  a  paper  by  Mr.  N.  W.  Akimoff,1  are  shown  in  Fig. 
113.  The  diagram  here  shown  represents  a  vertical  section  of 

J  Jour.  Amer.  Water  Works  Assoc.,  May,  1914. 


116  ELEMENTS  OF  HYDRAULICS 

a  current  flowing  vertically  downward  against  a  horizontal  cir- 
cular plate. 

The  stream  lines  S,  shown  by  the  full  lines  in  the  figure,  are 
curves  of  the  third  degree,  possessing  the  property  that  the 
volumes  of  the  cylinders  inscribed  in  the  surface  of  revolution 
generated  by  each  stream  line  are  equal.  For  instance,  the 
volume  of  the  circular  cylinder  shown  in  section  by  AA'BB'  is 
equal  to  that  of  the  cylinder  CC'DD',  etc.  It  may  also  be  noted 
that  the  size  of  the  plate  does  not  affect  the  general  shape  or 
properties  of  the  curves  shown  in  the  diagram. 

The  surfaces  of  equal  velocity  are  ellipsoids  of  revolution 
having  the  center  of  the  plate  0  as  center,  and  are  shown  in  sec- 
tion in  the  figure  by  the  ellipses  marked  EV.  In  general,  each 
of  these  ellipses  intersects  any  stream  line  in  two  points,  such 
as  F  and  G.  Therefore  somewhere  between  F  and  G  there  must 
be  a  point  of  minimum  velocity,  this  being  obviously  the  point 
of  contact  of  the  corresponding  ellipse  with  the  stream  line. 
The  locus  of  these  points  of  minimum  velocity  is  a  straight  line 
OH  in  section,  inclined  to  the  plate  at  an  angle  of  approxi- 
mately 20°.  The  surface  of  minimum  velocity  is  therefore  a 
cone  of  revolution  with  center  at  0,  of  which  OH  is  an  element. 

The  surfaces  of  equal  pressure  are  also  ellipsoids  of  revolution 
with  common  center  below  0,  and  are  shown  in  section  by  the 
ellipses  marked  PP  in  the  figure.  The  surface  of  maximum  pres- 
sure is  a  hyperboloid  of  revolution  of  one  sheet,  shown  in  section 
by  the  hyperbola  YOY. 

It  should  be  especially  noted  that  the  cone  of  minimum  ve- 
locity is  distinct  from  the  hyperboloid  of  maximum  pressure  so 
that  in  this  case  minimum  velocity  does  not  necessarily  imply 
maximum  pressure,  as  might  be  assumed  from  a  careless  applica- 
tion of  Bernoulli's  theorem. 

This  analysis  shows  the  reason  for  the  wide  variation  in  the 
results  obtained  by  different  experimenters  with  the  Pitot  tube, 
and  makes  it  plain  that  they  will  continue  to  differ  until  the  hy- 
draulic principles  underlying  the  action  of  the  impact  tube  are 
generally  recognized  and  taken  into  account. 

Construction  and  Calibration  of  Pitot  Tubes. — The  impact  end 
of  a  Pitot  tube  is  usually  drawn  to  a  fine  point  with  a  very  small 
orifice,  whereas  the  vertical  arm  is  given  a  much  larger  diame- 
ter in  order  to  avoid  the  effect  of  capillarity.  The  tubes  used 
by  Darcy  had  an  orifice  about  0.06  in.  in  diameter  which  was 


HYDROKINETICS  117 

enlarged  in  the  vertical  arm  to  an  inside  diameter  of  about  0.4  in. 
In  his  well-known  experiments  for  determining  the  velocity  of 
fire  streams  (Art.  21),  Freeman  used  for  the  mouthpiece  of  his 
impact  tube  the  tip  of  a  stylographic  pen,  having  an  aperture 
0.006  in.  in  diameter.  With  this  apparatus  and  for  the  high 
velocities  used  in  the  tests,  the  head  was  found  to  be  almost 

v2 
exactly  equal  to  ^  corresponding  to  a  value  of  m  =  1.0  in  the 

formula  v  =  m  ^2gh. 

It  is  also  important  that  the  impact  arm  should  be  long  enough 
so  that  its  orifice  is  clear  of  the  standing  wave  produced  by  the 
current  flowing  against  the  vertical  arm.  The  cutwater  used  with 
some  forms  of  apparatus  (see  Fig.  Ill)  is  intended  to  eliminate 
this  effect  but  it  is  doubtful  just  how  far  it  accomplishes  its 
purpose. 

The  most  prolific  source  of  error  in  Pitot-tube  measurements 
is  in  the  calibration  of  the  apparatus.  The  fundamental  principle 
of  calibration  is  that  the  tube  must  be  calibrated  under  the  same 
conditions  as  those  for  which  it  is  to  be  used.  Thus  it  has  been 
shown  in  Art.  16  that  flow  below  the  critical  velocity  follows  an 
entirely  different  law  from  that  above  this  velocity.  Flow  in  a 
pipe  under  pressure  is  also  essentially  different  from  flow  in  an 
open  channel. 

Du  Buat's  Paradox. — Furthermore,  the  method  of  calibration 
is  of  especial  importance.  This  is  apparent  from  the  well-known 
hydraulic  principle  known  as  Du  Buat's  paradox.  By  experiment 
Du  Buat  has  proved  that  the  resistance,  or  pressure,  offered  by  a 
body  moving  with  a  velocity  v  through  a  stationary  liquid  is 
quite  different  from  that  due  to  the  liquid  flowing  with  the  same 
velocity  v  past  a  stationary  object.  '  The  pressure  of  the  moving 
liquid  on  the  stationary  object  was  found  by  him  to  be  greater 
than  the  resistance  experienced  by  the  moving  object  in  a  station- 
ary liquid  in  the  ratio  of  13  to  10.  All  methods  of  calibration 
which  depend  on  towing  the  instrument  through  a  liquid  at  rest 
therefore  necessarily  lead  to  erroneous  and  misleading  results. 

Since  the  Pitot  tube  is  so  widely  used  for  measuring  velocity 
of  flow,  its  construction  and  calibration  should  be  standardized, 
so  that  results  obtained  by  different  experimenters  may  be  sub- 
ject to  comparison,  and  utilized  for  a  more  accurate  and  scientific 
construction  of  the  instrument. 


118 


ELEMENTS  OF  HYDRAULICS 


28.  NON-UNIFORM  FLOW;  BACKWATER 

Energy  Equation  for  Stream  of  Variable  Depth. — In  Art.  24 
it  is  shown  that  steady,  uniform  flow  implies  constant  slope  and 
cross  section  of  the  channel,  in  which  case  the  velocity  and  depth 
of  water  are  also  constant.  In  the  case  of  natural  channels, 
however,  the  cross  section  is  never  uniform  but  varies  from  point 
to  point,  in  consequence  of  which  the  flow  is  non-uniform,  both 
the  velocity  and  hydraulic  radius  changing  with  the  depth. 

For  a  stream  with  free  upper  surface,  the  hydraulic  gradient 
coincides  with  the  water  surface,  and  consequently  the  drop  in 
the  water  surface  in  any  distance  I  measures  the  head  lost  in  this 
distance. 


FIG.  114. 

Now  consider  two  adjacent  cross  sections  of  a  stream  of  vari- 
able depth  (Fig.  114).     Then  by  Bernoulli's  theorem  we  have 

Pa      .     *>12   _  pa  V22 


where  zif  22  denote  the  elevations  of  the  surface  at  the  two 
sections  considered  above  a  horizontal  datum  plane,  and  hi  is  the 
frictional  head  lost  in  the  given  distance  I.  Since  the  atmospheric 
pressure  pa  is  constant,  this  expression  simplifies  into 


(22  -  21)  + 


+  hi  =  0. 


An  expression  for  the  lost  head  hi  in  terms  of  the  average  velocity 
v  may  be  obtained  from  Chezy's  formula,  namely, 

v  =  C  V™, 
for  by  squaring  and  transposing,  it  becomes 


C2r 


HYDROKINETICS 


119 


and  since  y  =  s,  these  two  relations  combine  to  give 

to« 

fe  =  cv" 

Substituting  this  value  of  hi  in  the  above  expression  it  takes  the 
form 


(22  + 


(70) 


Differential   Equation   of   Surface   Profile.1 — To   obtain   the 
equation  of  a  longitudinal  profile  of  the  surface,  or  surface  curve 


FIG.  115. 

as  it  is  called,  let  the  distance  between  the  two  cross  sections 
considered  be  infinitesimal,  and  denote  this  distance  by  dx. 
Then  the  difference  in  elevation  z  will  also  be  infinitesimal,  that  is, 
22  —  Zi  =  —  dz,  and  also  v22  —  Vi2  =  d(v2)  =  2vdv,  where  d(vz) 
is  the  ordinary  differential  notation  indicating  an  infinitesimal 
change  in  v2.  Substituting  these  differential  values  in  Eq. 
(70),  it  becomes 

,   vdv   ,   v2dx  ,    x 

-  dz  +  —  +  TJ^T  =  o,  (71) 

which  is  therefore  the  differential  equation  of  the  surface  curve. 
Now  take  the  origin  on  the  down-stream  side  of  the  section  at  a 
distance  x  from  it,  and  suppose  the  adjacent  section  to  be  at  a 
distance  x  +  dx  from  the  origin,  as  indicated  in  Fig.  115.  If  s 
denotes  the  slope  of  the  bed  of  the  stream  and  y  the  depth  of  the 

1  The  method  here  given  for  determining  the  equation  of  the  surface  curve 
is  substantially  that  followed  in  all  standard  texts  on  hydraulics.  For  exam- 
ple, see  Foppl,  Vorlesungen  iiber  Technische  Mechanik,  Bd.  1. 


>i 

120  ELEMENTS  OF  HYDRAULICS 

water,  the  fall  of  the  bed  m  the  distance  dx  is  sdx  and  the  change 
in  depth  is  dy.  The/  latter,  however,  is  negative  since  the  slope 
of  the  surface  is  less  than  that  of  the  bed,  so  that 

dz  =  sdx  —  dy. 

Also,  if  A  denotes  the  area  of  the  cross  section  with  abscissa  x, 
b  its  average  width,  and  Q  the  quantity  of  water  flowing  past  this 
section  per  unit  of  time,  then 

A  =  by,  dA  =  bdy, 

,  =  |   =  |,  *=-|Jf- 

Substituting  these  values  for  dz  and  dv  in  Eq.  (71),  it  becomes 

Q2dy    ,     Q2dx 
dy  -  sdx  -  ^8  +  jfc»  =  o.  (72) 


Back-water  Curve  for  Broad  Shallow  Stream.  —  The  case  of 
most  practical  importance  is  that  in  which  the  level  of  a  stream  is 
raised  by  means  of  a  dam  or  weir,  and  it  is  required  to  determine 
the  new  elevation  of  the  surface  back  of  the  dam  or  weir.  In 
this  case  let  h  denote  the  original  depth  of  the  stream  before  the 
weir  was  built,  or  the  depth  below  the  weir  after  it  is  constructed 
(Fig.  116).  Then  assuming  that  the  breadth  of  the  stream  and 
the  total  discharge  remains  unchanged,  we  have 

Q  =  vbh  =  C  Vr's  bh 

where  r1  denotes  the  hydraulic  radius  for  the  original  channel 
section.     Substituting  this    value  of  Q  in  Eq.   (72)  it  becomes 

C2r'sbWdy   ,   Wr'sWdx 
dy-sdx-      —^^T-+-l^c^-       °> 

which  simplifies  into 

2  '2 

(73) 


If  the  stream  is  assumed  to  be  shallow  in  comparison  with  its 
breadth,  this  expression  can  be  greatly  simplified.  Since  the 
expressions  for  the  cross  sections  and  hydraulic  radii  of  the  new 
channel  and  the  original  channel  are 


HYDROKINETICS 


121 


if  y  and  h  are  small  in  comparison  with  the  breadth  b,  the  values 
of  the  hydraulic  radii  are  approximately 

r  =  y,  r'  =  h. 

Inserting  these  values  of  r,  r'  and  A  in  Eq.  (73)  and  multiplying 

l*< 
through  by  r^,  it  becomes 


(74) 

which  is  the  required  differential    equation  of  the   backwater 
curve  for  the  special  case  considered. 


FIG.  116. 


Integration  of  Backwater  Function. — To  integrate  Eq.  (74) 


let 


r-  =  u.     Then  dy  =  hdu,  and  Eq.  (74)  becomes 


u 


3    _ 


g 


hdu  =    uz  — 


which  may  be  written  in  the  form 


u 


3    _ 


By  ordinary  division, 

u* 1  - 

n 

1  + 


du. 


g 


u*  -  1  w3  -  1  ' 

and  consequently  the  above  relation  becomes 


SI  i*  2  o  \          slii 

I  O    6   \         liM 

5*rr*»  +  (i-  —)^—i 


122  ELEMENTS  OF  HYDRAULICS 

Now  integrating  between  limits  x\  and  x2  for  x,  and  corresponding 
limits  ui  and  u2  for  u,  we  have 


The  values  of  the  first  two  integrals  are  simply 

•fxi  rui 

I    dx  =  Xi  —  Xz  =  I,         I     du  =  HI  —  uz, 

Jxi  Juz 

where  I  denotes  the  distance  between  the  cross  sections  con- 
sidered. For  convenience  let  the  remaining  integral  be  denoted 
by  <p  (u)  ;  that  is,  let 


C    du         1  ,      w2  +  u  +  1        1 

*c«>;  -  J  urm°  e  log  (u  -  i)«     vi 

Substituting  the  limits,  we  have  in  this  notation 


,        2u  +  1 

cot 


and  therefore  Eq.  (75)  becomes 


or,  since  T  =  u,  we  have  y\  =  hui,  y%  =  huz  and  consequently  this 
equation  may  be  written 

-  ^)l^(ui)  -  <^(u2)],       (76) 

g  / 

which  is  the  required  equation  of  the  backwater  curve. 

Values  of  Integral.  —  To  avoid  the  labor  of  calculating  the 
numerical  value  of  the  integral  <p(u)  in  any  given  case,  the  values 
of  the  integral  corresponding  to  values  of  u  ordinarily  occurring 
in  practice  are  tabulated  below:1 

1  Bresse,  Mecanique  Appliques,  Vol.  2.  Also  given  in  Williamson's 
Integral  Calculus;  in  Hoskin's  Text-book  on  Hydraulics,  and  elsewhere. 


HYDROKINETICS 


123 


Numerical  values  of  <?(u)  = 

/du 

1 

u'- 

u 

<p(u)      ||      u 

<p(u}    ||        u             <f>(u]              u 

<p(u) 

1.00 

oo 

1.10 

0.680 

1.30 

0.373 

1.65 

0.203 

1.01 

1.419 

1.12 

0.626 

1.32 

0.357 

1.70 

0.189 

1.02 

1.191 

1.14 

0.581 

1.34 

0.342 

1.80 

0.166 

1.03 

1.060 

1.16 

0.542 

1.36 

0.328 

1.90 

0.147 

1.04 

0.967 

1.18 

0.509 

1.38 

0.316 

2.00 

0.1318 

1.05 

0.896 

1.20 

0.480 

1.40 

0.304 

2.10 

0.1188 

1.06 

0.838 

1.22 

0.454 

1.45 

0.278 

2.20 

0.1074 

1.07 

0.790 

1.24 

0.431 

1.50 

0.255 

2.30 

0.0978 

1.08 

0.749 

1.26 

0.410 

1.55 

0.235 

2.40 

0.0894 

1.09 

0.713 

1.28 

0.390 

1.60 

0.218 

2.50 

0.0822 

Condition  for  Singularities  in  Backwater  Curve. — If  the  con- 
stant C  in  Chezy's  formula  has  the  numerical  value 

C  =  90  ft.*  sec.-1, 

which  may  be  assumed  as  its  average  value  under  ordinary 
conditions,  then 

C2 

—  =  250,  approximately. 

Consequently,  when  the  slope  s  has  the  value 


250' 


then 


and  Eq.  (76)  for  the  backwater  curve  reduces  to 

2/i  =  2/2  +  si 
If  the  fall  is  steeper  than        '  there  are  singularities  in  the 


backwater  curve.     This  is  apparent  from  Eq.  (72);  for,  writing 
it  in  the  form 


dy 
dx 


1  - 


gA* 


124  ELEMENTS  OF  HYDRAULICS, 

it  is  evident  that  the  curve  will  have  a  vertical  tangent,  that  is, 

r     =  oo  }  if  the  denominator  of  this  fraction  is  zero,  that  is,  if 


But  since  Q  =  Av  =  byC  ^ys,  this  condition  may  be  written 

i  _  b2y*C2ysb 
gbV 

or,  cancelling  common  factors, 

1-^-0 
9 

which  is  identical  with  the  condition  previously  obtained.  Con- 
sequently, if  s  >  250  approximately,  there  will  be  a  sudden  ver- 
tical drop  in  the  backwater  curve,  whereas  if  s  is  less  than  this 
critical  value,  -r-  is  always  negative  and  the  curve  contains  no 

such  singularity. 

Numerical  Application.  —  To  illustrate  the  application  of  Eq. 
(76),  suppose  that  on  a  certain  river  it  is  proposed  to  create 
a  waterfall  for  a  hydraulic  power  station  by  building  a  dam. 
Before  proceeding  with  the  construction  it  is  required  to  know 
among  other  things  whether  the  water  can  be  raised  5  ft.  at  the 
dam  without  interfering  with  the  operation  of  an  existing  power 
plant  7  miles  up  stream. 

Suppose  that  the  stream  at  the  proposed  site  is  10  ft.  deep  and 
500  ft.  wide,  the  slope  of  its  bed  is  2  ft.  per  mile,  and  the  maximum 
annual  discharge  is  25,000  cu.  ft.  per  second.  Then  its  mean 
velocity  is 

25,000  , 

v  "=  500  X  10  =  5  ft'  per  sec"> 

the  hydraulic  radius  for  the  given  section  is 

500  X  10 
500  +  20 

and  the  value  of  the  constant  C  in  Chezy's  formula  is 

5 


=  82.9. 


5280 


HYDROKINETICS  125 

First,  let  it  be  required  to  find  how  far  up  stream  the  increase 
in  depth  will  amount  to  2  ft.     From  the  given  data, 

h  =  10,    s  =  g|g,    C  =  82.9,    yi  --=  15,    2/2  =  12, 
15  12 

Ul  ==  10  =:  L5j    U2  =  10  =  L2j 

and  from  the  table  for  <p(u)  on  page  123, 

<P(UI)  =  0.255,      <p(u2)  =  0.480. 
Substituting  these   numerical  values  in  Eq.    (76)   it  becomes 


(0.255  - 


0.480)  =  13,385  ft.  =  2-1/2  miles. 


Second,  to  find  the  increase  in  depth  at  the  existing  plant  7 
miles  up  stream,  the  given  data  are 

h  =  10,    s  =          ,    C  =  82.9,    2/1  =  15,    Wl  =  1.5, 


2/2)  - 

2    1 

5- 

8 

2'9  5280 

*~ 

32.2      J 
[0.25 

<P(UI)  =  0.255,    I  =  7(5280), 

the  unknown  quantities  being  yz  and  ^(^2).     Substituting  the 
known  quantities  in  Eq.  (76),  it  becomes 


36,960  -  ^-(15 


which  reduces  to 

9.2?(w2)  =  2/2  +  1-346. 

This  equation  can  best  be  solved  by  trial  by  -assuming  a  value  for 
2/2,  taking  the  corresponding  value  of  <p(uz)  from  the  table,  and 
substituting  these  trial  values  in  the  equation.  Thus,  assuming 
the  rise  to  be  0.2  ft.,  we  have 

10  2 
1/2  =  10.2,    u2  =  --  =  1.02,    9(u,)  =  1.191, 


and  the  left  member  of  the  equation  becomes 
9.2<p(u2)  =  10.957, 


126 


ELEMENTS  OF  HYDRAULICS 


whereas  the  right  member  is 

2/2  +  1.346  =  11.546. 
Similarly,  assuming  the  rise  to  be  0.1  ft.,  we  have 


=  10.1,^2  =  ~~  =  1.01, 

and  the  left  member  now  becomes 

=  13.055, 


=  1.419, 


whereas  the  right  member  is 

2/2  +  1.346  =  11.446. 

It  is  evident,  therefore,  that  the  actual  rise  in  surface  elevation  at 
the  given  point  is  slightly  less  than  0.2  ft. 

APPLICATIONS 

61.  A  device  used  by  Prony  for  measuring  discharge  consists 
of  a  fixed  tank  A  (Fig.  117)  containing  water,  in  which  floats  a 

cylinder  C  which  carries  a  second 
tank  B.  Water  flows  through  the 
opening  D  from  A  into  B.  Show  that 
the  head  on  the  opening  D,  and  con- 
sequently the  velocity  of  flow 
through  this  opening,  remains  con- 
stant. (Wittenbauer.) 

52.  A  cylindrical  tank  of  6  ft.  in- 
side diameter  and  10  ft.  high  con- 
tains 8  ft.  of  water.  An  orifice  2  in. 
in  diameter  is  opened  in  the  bottom, 
and  it  is  found  that  the  water  level  is 

lowered  21  in.  in  3  minutes.  Calculate  the  coefficient  of  discharge. 
53.  Water  flows  through  a  circular  sharp-edged  orifice  1/2  in. 
in  diameter  in  the  side  of  a  tank,  the  head  on  the  center  of  the 
opening  being  6  ft.  A  ring  slightly  larger  than  the  jet  is  held  so 
that  the  jet  passes  through  it,  and  it  is  then  found  that  the  center 
of  the  ring  is  3  ft.  distant  from  the  orifice  horizontally ,  and  8.23  ft. 
below  it.  In  5  minutes  the  weight  of  water  discharged  is  301  Ib. 
Calculate  the  coefficients  of  velocity,  contraction  and  discharge 
for  this  orifice. 

Note. — This  is  an  interesting  method  of  determining  the  coeffi- 
cients by  experiment  but  is  not  very  accurate. 


FIG.  117. 


HYDROKINETICS  127 

If  the  velocity  of  the  jet  at  exit  is  denoted  by  v,  its  abscissa  x 
after  t  seconds  will  be  approximately 

x  =  vt 

and  the  ordinate  of  the  same  point,  considering  the  water  as  a 
freely  falling  body,  will  be 


Eliminating  t  between  these  two  relations,  the  equation  of  the 
path  followed  by  the  jet  is  found  to  be 


g 

which  represents  a  parabola  with  axis  vertical  and  vertex  at  the 
orifice.  Having  found  the  actual  velocity  v  from  this  equation, 
the  velocity  coefficient  is  obtained  from  the  relation 

4) 

c,  = 


The  efflux  coefficient  K  is  then  calculated  from  the  measured  dis- 
charge Q  from  the  relation 

Q  =  KAv, 
and  the  contraction  coefficient  from 

c  ---• 

v/«  — •  /m 

U  v 

54.  Find  the  velocity  with  which  water  will  flow  through  a 
hole  in  a  steam  boiler  shell  at  a  point  2  ft.  below  the  surface  of  the 
water  when  the  steam  pressure  gage  indicates  70  Ib.  per  square 
inch. 

55.  A  reservoir  having  a  superficial  area  of  0.5  sq.  mile  has  an 
outlet  through  a  rectangular  notch  weir  8  ft.  long.     If  the  head 
on  the  crest  when  the  weir  is  opened  is  2.5  ft.,  how  long  will  it 
take  to  lower  the  level  of  the  reservoir  1  ft.? 

56.  A  rectangular  notch  weir  12  ft.  long  has  a  head  of  15  in.  of 
water  on  the  crest.     The  cross-sectional  area  of  the  approach 
channel  is  50  sq.  ft.     Calculate  the  flow. 

57.  A  suppressed  weir  6  ft.  long  has  its  crest  3  ft.  above  the 
bottom  of  the  channel,  and  the  head  on  the  crest  is  18  in.     Com- 
pute the  discharge. 


128  ELEMENTS  OF  HYDRAULICS 

58.  A  lock  chamber  500  ft.  long  and  110  ft.  wide  is  emptied 
through  a  submerged  opening  6  ft.  long  by  3  ft.  high,  having  a 
coefficient  of  discharge  of  0.58.     If  the  depth  of  water  on  the 
center  of  the  opening  is  initially  30  ft.  on  the  inside  and  8  ft.  on 
the  outside,  find  how  long  it  will  take  to  lower  the  water  in  the 
lock  to  the  outside  level. 

59.  A  hemisphere  filled  with  water  has  a  small  orifice  of  area  A 
at  its  lowest  point.     Calculate  the  time  required  for  it  to  empty. 

Note. — If  x  denotes  the  depth  of  water  at  any  instant,  the 
areaX  of  the  water  surface  is  X  =  ir(2rx  —  x2),  where  r  denotes 
the  radius  of  the  hemisphere.  The  time  required  to  empty  the 
hemisphere  is  therefore 

rXdx        14      irr^ 


KA  V20J0    Vz         15  KA  V20 

60.  A  tank  10  ft.  square  and  12  ft.  deep  is  filled  with  water.     A 
sharp-edged  circular  orifice  3  in.  in  diameter  is  then  opened  in  the 
bottom.     How  long  will  it  take  to  empty  the  tank  through  this 
opening? 

61.  Compute  the  discharge  through  a  Borda  mouthpiece  1.5 
in.  in  diameter  under  a  head  of  12  ft.,  and  determine  the  Joss  of 
head  in  feet. 

62.  Compute  the  discharge  through  a  re-entrant  short  tube  2 
in.  in  diameter  under  a  head  of  20  ft.,  and  determine  the  loss  of 
head  in  feet. 

63.  Compute  the  discharge  through  a  standard  short  tube  of 
1.75  in.  inside  diameter  under  a  head  of  6  ft.,  and  also  find  the 
negative  pressure  head  at  the  most  contracted  section  of  the  vein. 

64.  Find  the  discharge  in  gallons  per  minute  through  a  1.5  in. 
smooth  fire  nozzle  attached  to  a  2.5  in.  play  pipe  under  a  pres- 
sure at  base  of  nozzle  of  90  Ib.  per  square  inch. 

65.  Water  flows  through  a  6-in.  horizontal  pipe  at  200  ft.  per 
minute  under  a  pressure  of  30  Ib.  per  square  inch.     If  the  pipe 
gradually  tapers  to  4  in.  diameter,  find  the  pressure  at  this  point. 

66.  A  12-in.  horizontal  pipe  gradually  tapers  to  a  diameter  of 
6  in.     If  the  flow  is  50,000  gal.  per  hour,  calculate  the  difference 
in  pressure  at  two  sections  having  these  diameters. 

67.  A  Venturi  meter  in  an  18-in.  main  tapers  to  6  in.  at  the 
throat,  and  the  difference  in  pressure  in  main  and  throat  is 
equivalent  to  11  in.  of  mercury.     Find  the  discharge  in  gallons 
per  minute. 


HYDROKINETICS  129 

68.  The  Ashokan  Venturi  meter  on  the  line  of  the  Catskill 
Aqueduct  is  7  ft.  9  in.  inside  diameter  at  the  throat,  the  diameter 
of  the  main  being  17  ft.  6  in.  (see  Fig.  71,  page  69).     Find  the 
difference  in  pressure  between  main  and  throat  for  the  estimated 
daily  flow  of  500,000,000  gal. 

69.  The  velocity  of  flow  in  a  water  main  4  ft.  in  diameter  is  3.5 
ft.  per  second.     Assuming  the  coefficient  of  friction  to  be  0.0216, 
find  the  frictional  head  lost  in  feet  per  mile. 

70.  Two  cylindrical  tanks  each  8  ft.  in  diameter  are  connected 
near  the  bottom  by  a  2-in.  horizontal  pipe  25  ft.  long.     If  the 
water  level  in  one  tank  is  initially  12  ft.  and  in  the  other  3  ft. 
above  the  center  line  of  the  pipe,  find  how  long  it  will  take  for  the 
water  to  reach  the  same  level  in  both  tanks. 

71.  Find  the  frictional  head  lost  in  a  pipe  2  ft.  in  diameter  and 
5  miles  long  which  discharges  200,000  gal.  per  hour,  assuming  the 
coefficient  of  friction  to  be  0.024. 

72.  Find  the  required  diameter  for  a  cast-iron  pipe  10  miles 
long  to  discharge  60,000  gal.  per  hour  under  a  head  of  200  ft. 

73.  A  house  service  pipe  is  required  to  supply  4000  gal.  per 
hour  through  a  1.5-in.  pipe  and  a  1-in.  tap.     The  total  length  of 
the  service  pipe  is  74  ft.,  including  the  tap  which  is  1.5  ft.  long. 
Find  the  total  pressure  required  in  the  main. 

Solution.  —  In  the  solution  of  water-supply  problems  of  this  type, 
it  is  recommended  by  W.  P.  Gerhard1  that  the  following  formulas 
be  used. 


Head  lost  in  tap,  hi  =  0.024  -      -> 

v2 
Head  lost  at  entrance,  hi  =  0.62  H~» 

Head  lost  at  stopcock  =1/2  head  lost  in  tap, 
Head  lost  in  pipe  by  Prony's  formula, 


where,  in  this  last  formula, 

d  =  diameter  of  pipe  in  inches, 
H  =  head  in  feet, 
L  =  length  in  yards, 
G  =  discharge  in  U.  S.  gal.  per  minute. 

1  Discharge  of  Water  through  Street   Taps  and  House  Service  Pipes, 
Cassier's  Mag.,  Nov.,  1905. 


130  ELEMENTS  OF  HYDRAULICS 

Using  these  formulas  we  obtain  in  the  present  case  the  following 
numerical  results  : 

Pressure  lost  in  tap  =    2. 24  Ib.  per  square  inch. 

Pressure  lost  at  stopcock  =     1 . 12  Ib.  per  square  inch. 

Pressure  lost  at  entrance  =    3 . 08  Ib.  per  square  inch. 

Pressure  lost  in  72.5  ft.  of  1-in.  pipe  =  17.53  Ib.  per  square  inch. 
Total  pressure  required  in  main         =  23 . 97  Ib.  per  square  inch. 

74.  A  building  is  to  be  supplied  with  2500  gal.  of  water  per 
hour  through  180  ft.  of  service  pipe  at  a  pressure  at  the  building 
line  of  15  Ib.  per  square  inch.     The  pressure  in  the  main  is  35  Ib. 
per  square  inch.     Find  the  required  size  of  service  pipes  and 
taps. 

Solution. — The  total  drop  in  pressure  in  this  case  is  20  Ib.  per 
square  inch.  Therefore  using  the  formulas  given  in  the  pre- 
ceding problem  and  assuming  different  sizes  of  service  pipes,  the 
results  are  as  follows: 

One  1.25-in.  full-size  pipe  180  ft.  long  discharges  1715  gal.  per 
hour. 

Two  1-in.  full-size  pipes  discharge  together  1920  gal.  per  hour. 

Two  1.25-in.  pipes  with  5/8-in.  taps  discharge  together  2880 
gal.  pe*r  hour. 

One  1.5-in.  pipe  with  1-in.  tap  discharges  2519  gal.  per  hour. 

The  last  has  sufficient  capacity  and  is  cheapest  to  install,  and 
is  therefore  the  one  to  be  chosen. 

75.  A  pipe  1  ft.  in  diameter  connects  two  reservoirs  3  miles 
apart  and  has  a  slope  of  1  per  cent.     Assuming  the  coefficient 
of  friction  as  0.024,  find  the  discharge  and  the  slope  of  the  hydrau- 
lic gradient  when  the  water  stands  30  ft.  above  the  inlet  end  and 
10  ft.  above  the  outlet  end. 

76.  Two  reservoirs  5  miles  apart  are  connected  by  a  pipe  line 
1  ft.  in  diameter,  the  difference  in  water  level  of  the  two  reservoirs 
being  40  ft.     Assuming  the  value  of  Chezy's  constant  in  feet 
and  second  units  to  be  125,  find  the  discharge  in  gallons  per 
hour. 

77.  A  12-in.  main  5000  ft.  long  divides  into  three  other  mains, 
one  6  in.  in  diameter  and  6000  ft.  long,  one  10  in.  in  diameter  and 
7000  ft.  long,  and  one  8  in.  in  diameter  and  4000  ft.  long.     The 
total  static  head  lost  in  each  line  between  reservoir  and  outlet  is 
the  same  and  equal  to  100  ft.     Find  the  discharge  in  gallons  per 
24  hours  at  each  of  the  three  outlets. 


HYDROKINETICS 


131 


Solution.  —  The  head  lost  in  friction  in  the  length  I  is  given  by 
the  relation 


and  the  discharge  by 


Eliminating  v  between  these  relations,  we  have 

16/ZQ2 


Q  = 

10S 

ht 


rd2 


whence 


Q  = 


8/7 


Assuming/  =  0.02  and  /  =  1000  ft.,  the  discharge  Q  for  pipes  of 
various  sizes  in  terms  of  the  head  lost  per  1000  ft.  is  given  by 
the  following  relations : 


Diameter  of  pipe  in 
inches 

Discharge  in  gal.  per  24  hours  in  terms  of  head 
lost  per  1000  ft. 

4 

Q  =          58,430  Vhi 

6 

161,OOOVfe 

8 

330,500Vfo 

10 

=       577,500Vfo 

12 

=       911,000V/?! 

16 

=    l,870,OOOVfo 

20 

=    3,266,OOOVfo 

24 

=    5,147,OOOVfo 

30 

=    9,002,OOOVfo 

36 

=  14,200,OOOVfo 

48 

=  29,  150,000  Vhi 

56 

=  42,850,OOOVfo 

60 

=  50,920,OOOV/ii 

66  ' 

=  64,600,OOOVfe 

72 

=  80,320,000  Vhi 

In  the  present 
noted  by  Q  with 

Also  if  h  with  the 
pipe  per  1000  ft., 


case  let  the  flow  in  gallons  per  24  hours  be  de- 
a  subscript  indicating  the  size  of  pipe.     Then 

612  =  Qe  +  Qs  +  Qio. 

proper  subscript  denotes  the  head  lost  in  each 
we  have  from  the  above  relations 


Q 


=  911,000 
=  161,000 
=  330,500 
=  577,500 


132 


ELEMENTS  OF  HYDRAULICS 


and  since  from  the  conditions  of  the  problem  the  head  lost  in 
each  line  amounts  to  100  ft.,  we  also  have  the  relations 

5Ai2  +  6A6  =  100, 
5/*i2  +  7hi0  =  100, 
5&i2  +  4/*8  =  100. 


From  these  relations  we  find 


100  - 


and  substituting  these  values  in  the  first  equation,  the  result  is 
F  E  D  C  B  A 


1 

9. 

^ 

Q, 

«, 

1 

M   N  R 
FIG.  118. 


161,000  V/i6  +  330,500 


+  577,500 


=  91 1,000  J 


100  - 


whence  ^6  =  7.52  and  consequently 

h8  =  12.28;   hlo  =  6.446;   h12  =  10.976. 

Substituting  these  values  of  h  in  the  formulas  for  discharge,  the 
results  are 

Q6  =     444,360  gal.  per  24  hr. 

Q8  =  1,110,480  gal.  per  241ir. 

Qio  =  1,465,700  gal.  per  24  hr. 

Oe  +  Qs  +  Qio  =  3,020,540  gal.  per  24  hr. 

The  actual  calculated  value  of  Qi%  is 

Qi2  =  3,015,400  gal.  per  24  hr., 


HYDROKINETICS  133 

the  discrepancy  between  these  results  being  due  to  slight  inac- 
curacy in  extracting  the  square  roots. 

78.  A  pipe  of  constant  diameter  d  discharges  through  a  number 
of  laterals,  each  of  area  A  and  spaced  at  equal  distances  I  apart 
(e.g.,  street  main  and  house  service  connections).  Find  the  rela- 
tion between  the  volume  of  flow  in  three  successive  segments  of 
the  main1  (Fig.  118). 

Solution.  —  The  discharge  at  A  is 

Oi  =  KA 


where  vi  denotes  the  velocity  of  flow  at  this  point.     Also  the  head 
lost  in  friction  in  the  segment  AB  is 

l 


f(\   Vl 
=f(d)    2 


and  consequently 

where 

8/7 

^    =  2^75* 

At  B  the  pressure  head  is  h  +  hi  and  the  discharge  is 

Q2  -  Q,  =  KA 

Similarly,  for  the  discharge  at  C  and  D  we  obtain  the  relations 
O,  -  Q2  =  KA 


Qt  —  Q3  =  KA  V20  V/i  +  a(Qi< 
whence  by  elimination 

where 

62  = 


The  general  relation  is  therefore 

(Q»    ~   Qn-l)2   ~    (Qn-l    -   Qn-t)2    = 

The  following  geometrical  construction  may  be  used  for  deter- 
mining Qn.     Determine  an  angle  0  such  that 

6  =  tan  6 
1  J.  P.  Frizell,  Jour.  Franklin  Inst.,  1878. 


134  ELEMENTS  OF  HYDRAULICS 

and  lay  off  Qn-i  and  Qn_2  on  a  straight  line  so  that  OM  =  Qn_2 
and  ON  =  Qn--\  as  shown  in  Fig.  118.  At  N  erect  a  perpendicular 
NP  to  ON,  and  then  lay  off  NR  =  MP.  Then  OR  =  Qn. 

79.  A  reservoir  discharges  through  a  pipe  line  made  up  of  pipes 
of  different  sizes,  the  first  section  being  4000  ft.  of  24-in.  pipe, 
followed  by  5000  ft.  of  20-in.  pipe,  6000  ft.  of  16-in.  pipe  and  7000 
ft.  of  12-in.  pipe.  The  outlet  is  100  ft.  below  the  level  of  the 
reservoir.  Find  the  discharge  in  gallons  per  24  hours. 

Solution.  —  Using  the  same  notation  as  in  Problem  77,  we  have 
in  the  present  case 

4/>24  +  5/*2o  +  6Ai6  +  7h12  =  100. 
Also,  since 


the  loss  in  head  per  1000  ft.  varies  inversely  as  the  fifth  power 
of  the  diameter,  and  consequently 

24 


=  ^4=    7.594A24; 


Solving  these  three  equations  simultaneously  with  the  first  one, 
the  results  are 

^24  =  0.35;  /*2o  =  0.871;  A16  =  2.658;  h12  =  11.20. 
As  a  check  on  the  correctness  of  these  results  we  have 

4X    0.35     =     1.400 

5  X    0.871  =    4.355 

6  X    2.658  =  15.948 

7  X  11.20    =  78.400 

100.103 

The  discharge  may  be  found  from  the  formulas  given  in  Problem 
77,  the  results  being  as  follows: 

Qi2  =  Qie  =  Q2o  =  Q24  =  3,046,000  gal.  per  24  hr. 

80.  A  pipe  AB,  1000  ft.  in  length,  divides   at  B  into  two 
pipes,  BC  which  is  600  ft.  long  and  BD  which  is  900  ft.  long.     The 


HYDROKINETICS 


135 


fall  for  AB  is  25  ft.,  for  BC  is  10  ft.  and  for  BD  is  20  ft.  Find  the 
required  diameters  of  the  three  pipes  to  deliver  500  gal.  per 
minute  at  C  and  300  gal.  per  minute  at  D. 

81.  A  reservoir  empties  through  a  pipe  AB  (Fig.  119)  which 
branches  at  B  into  two  pipes  BC  and  BD,  one  of  which  discharges 


FIG.  119. 


20,000  gal.  per  hour  at  C  and  the  other  30,000  gal.  per  hour  at  D. 
The  lengths  of  the  pipes  are  AB  =  1200  ft.,  BC  =  900  ft.,  BD  = 
600  ft.,  and  the  depths  of  the  outlets  below  the  surface  of  the 
reservoir  are  hi  =  25  ft.,  A2  =  60  ft.  The  pipes  are  of  cast  iron, 


r 


FIG.  120. 


and  the  velocity  of  flow  in  A  B  is  to  be  3  ft.  per  second.     Calculate 

the  diameters  of  all  three,  and  the  velocity  of  flow  in  BC  and  BD. 

82.  Two  reservoirs  empty  through  pipes  which  unite  at  C 

(Fig.  120)  into  a  single  pipe  which  discharges  at  D.     The  lengths 


136 


ELEMENTS  OF  HYDRAULICS 


of  the  pipes  are  h  =  1500  ft.,  12  =  900  ft.  and  I  =  2400  ft.  The 
diameters  of  the  pipes  are  d±  =  6  in.,  d2  =  4  in.,  and  d  =  9  in., 
and  the  depths  of  the  outlet  below  the  levels  of  the  reservoirs 
are  hi  =  75  ft.,  A2  =  100  ft.  Find  the  velocity  of  flow  in  each 
pipe  and  the  total  discharge  in  gallons  per  hour. 

83.  A  water  main  3  ft.  in  diameter  divides  into  two  smaller 
mains  of  the  same  diameter  and  whose  combined  area  equals 
that  of  the  large  main.  If  the  velocity  of  flow  is  3  ft.  per  second, 
compare  the  heads  lost  per  mile  in  the  large  and  small  mains. 


Old     Croton. 


Aqua  Claudia. 
FIG.  121. — Comparison  of  ancient  and  modern  aqueducts. 

84.  The  head  on  a  fire  hydrant  is  300  ft.     Find  its  discharge  in 
gallons  per  minute  through  400  ft.  of  inferior  rubber-lined  cotton 
hose  2.5  in.  in  diameter  and  a  1.5-in.  smooth  nozzle. 

85.  What  head  is  required  at  a  fire  hydrant  to  discharge  250 
gal.  per  minute  through  a  1.25-in.  ring  nozzle  and  500  ft.  of 
2.5-in.  best  rubber-lined  cotton  hose? 


HYDROKINETICS 


137 


86.  A  fire  stream  is  delivered  through  100  ft.  of  2-in.  rubber- 
lined  cotton  hose  and  a  nozzle  1-1/8  in.  in  diameter.     The  hy- 
drant pressure  is  75  Ib.  per  square  inch.     Find  pressure  at  nozzle, 
discharge  in  gallons  per  minute  and  height  of  effective  fire  stream. 

87.  Two  reservoirs  are  connected  by  a  siphon  16  in.  in  diameter 
and  50  ft.  long.     If  the  difference  in  level  in  the  reservoirs  is 
25  ft.,  calculate  the  discharge,  assuming  the  coefficient  of  pipe 
friction  to  be  0.02  and  considering  only  friction  losses. 

88.  A  cast-iron  pipe  2  ft.  in  diameter  has  a  longitudinal  slope 
of  1  in  2500.     If  the  depth  of  water  in  the  pipe  is  18  in.,  calculate 
the  discharge. 

89.  A  rectangular  flume  6  ft.  wide,  3  ft.  deep  and  1  mile  long 
is  constructed  of  unplaned  lumber  and  is  required  to  deliver  120 
cu.  ft.  per  second.     Determine  the  necessary  gradient  and  the 
total  head  lost. 

90.  The  Aqua  Claudia,  shown  in  Fig.  121,  was  one  of  the  nine 
principal  aqueducts  in  use  in  the  First  Century,  A.  D.,  for  supply- 
ing Rome  with  water.     The  lengths  and  capacities  of  these 
nine  aqueducts  were  as  follows: 


Name 

Date  of 
construc- 
tion 

P 

Length 
in 
feet 

Length 
in 
miles 

Discharge 
per  day 
in  cu.  ft. 

Alti- 
tude  of 
springs 
above 
sea 
level 
in  feet 

Level 
in 
Rome 
in  feet 

Aqua  Appia  
Anio  Vetus  
Aqua  Marcia.  .  . 
Aqua  Tepula  .  .  . 
Aqua  Julia  .  . 

312  B.  C. 
272-269 
144-140 
125 
33 

53,950 
209,000 
299,960 
58,200 
74,980 

10.2 
39.6 
56.8 
11.0 
14  2 

4,072,500 
9,814,200 
10,465,800 
993,000 
2,691,200 

98 
918 
1043 
495 
1148 

65 
157 
192 
199 
209 

Aqua  Virgo  
Aqua  Absietina. 

19 

67,900 
107,775 

12.9 
20.4 

5,587,700 
874,800 

79 
685 

65 
54 

Aqua  Claudia.  .  . 
Anio  Novus  .... 

38-52  A.  D. 
38-52  A.  D. 

225,570 
285,330 

42.7 
54.0 

7,390,800 
10,572,900 

1050 
1312 

221 
231 

The  construction  of  the  earliest  aqueducts  was  the  simplest,  most  of  them 
being  underground.  In  the  Aqua  Appia  only  300  ft.  were  above  ground, 
and  in  the  Anio  Vetus  only  1100  ft.  were  above  ground.  In  the  Aqua 
Marcia  7.5  miles  were  supported  on  arches;  in  the  Aqua  Claudia  10  miles 
were  on  arches,  and  in  the  Anio  Novus  9.5  miles  were  on  arches.  The  con- 
struction of  the  last  shows  the  greatest  engineering  skill,  as  it  follows  a 
winding  course,  at  certain  points  tunnelling  through  hills  and  at  others  cross- 
ing ravines  300  ft.  deep. 


138  ELEMENTS  OF'  HYDRAULICS 

The  cross  section  of  the  channels  (specus)  varied  at  different  points  of  the 
course,  that  of  the  largest,  the  Anio  Novus,  being  3  to  4  ft.  wide  and  9  ft. 
high  to  the  top,  which  was  of  pointed  shape.  The  channels  were  lined  with 
hard  cement  (opus  signinum)  containing  fragments  of  broken  brick.  The 
water  was  so  hard  that  it  was  necessary  to  clean  out  the  calcareous  deposits 
frequently,  and  for  this  purpose  shafts  or  openings  were  constructed  at 
intervals  of  240  ft. 

Filtering  and  settling  tanks  (pisdnce  limarioe,  or  "purgatories")  were 
constructed  on  the  line  of  the  aqueduct  just  outside  the  city,  and  within  the 
city  the  aqueducts  ended  in  huge  distributing  reservoirs  (Castella)  from  which 
the  water  was  conducted  to  smaller  reservoirs  for  distribution  to  the  various 
baths  and  fountains. 

Supposing  the  population  of  Rome  and  suburbs  to  have  then  numbered 
one  million,  there  was  a  daily  water  supply  of  nearly  400  gal.  per  capita. 
Modern  Rome  with  a  population  of  half  a  million  has  a  supply  of  about  200 
gal.  per  capita.  The  volume  of  water  may  also  be  compared  with  that  of  the 
Tiber  which  discharges  342,395,000  gal.  per  day,  whereas  in  the  First  Cen- 
tury, A.  D.,  the  aqueducts  carried  not  less  than  392,422,500  gal.  per  day, 
which  by  the  Fourth  Century  had  been  increased  by  additional  supplies  to 
461,628,200  gal.  per  day. 
i 

Assuming  that  the  Aqua  Claudia  had  an  average  width  of  3  ft. 
with  6  ft.  depth  of  water,  and  that  the  grade  was  uniform  and  the 
difference  in  head  lost  in  friction,  calculate  from  the  values  tabu- 
lated above  the  velocity  of  flow  and  Chezy's  constant  C  in  the 
formula  v  =  C  Vrs. 

91.  A  channel  of  trapezoidal  section  with  side  slopes  of  two 
horizontal  to  one  vertical  is  required  to  discharge  100  cu.  ft.  per 
second  with  a  velocity  of  flow  of  3  ft.  per  second.     Assuming 
Chezy's  constant  as  115,  compute  the  required  bottom  width  of 
channel  and  its  longitudinal  slope. 

92.  A  channel  of  trapezoidal  cross  section  has  a  bottom  width 
of  25  ft.  and  side  slopes  of  1  to  1.     If  the  depth  of  water  is  6  ft. 
and  the  longitudinal  inclination  of  the  bed  is  1  in  5000,  find  the 
discharge,  assuming  the  coefficient  of  roughness,  n,  in  Kutter's 
formula  to  be  0.02. 

93.  A  channel  of  rectangular  section  has  a  bottom  width  of 
20  ft.,  depth  of  water  6  ft.  and  longitudinal  slope  of  1  in  1000. 
Calculate  the  discharge,  assuming  the  coefficient  of  roughness, 
n,  in  Kutter's  formula  to  be  0.01. 

94.  A  reservoir  A  supplies  another  reservoir  B  with  400  cu.  ft. 
of  water  per  second  through  a  ditch  of  trapezoidal  section,  with 
earth  banks,  5  miles  long.     To  avoid  erosion,  the  flow  in  this 
channel  must  not  exceed  2  ft.  per  second. 


HYDROKINETICS 


139 


From  reservoir  B  the  water  flows  to  three  other  reservoirs, 
C,  D,  E.  From  B  to  C  the  channel  is  to  be  rectangular  in  section 
and  4  miles  long,  constructed  of  unplaned  lumber,  with  a  10-ft. 
fall  and  a  discharge  of  150  cu.  ft.  per  second. 

From  B  to  D  the  channel  is  to  be  5  miles  long,  semicircular  in 
section  and  constructed  of  concrete,  with  12-ft.  fall  and  a  discharge 
of  120  cu.  ft.  per  second. 

From  B  to  E  the  channel  is  to  be  3  miles  long,  rectangular  in 
section  and  constructed  of  rubble  masonry,  with  15  ft.  fall  and  a 
discharge  of  130  cu.  ft.  per  second. 

Find  the  proper  dimensions  for  each  channel  section. 

95.  The  flow  through  a  circular  pipe  when  completely  filled  is 
25  cu.  ft.  per  second  at  a  velocity  of  9  ft.  per  second.  How  much 


0.2        0.4 


0.6 


0.8 


1.0 


1.2 


[  Ratio  of  Depth  to  Diameter  | 

^ 

s^ 

Y 

1.0 

0.8 
O.G 
0.4 
0.2 
0 

\ 

/ 

I 

\ 

^ 

^ 

i 

\ 

Qx 

/ 

A 

/ 

^ 

/ 

. 

X 

/ 

/ 

' 

/ 

^ 

/ 

/ 

/ 

/ 

r 

/ 

/ 

^ 

S 

f 

^ 

^ 

^ 

0  0.2          0.4          0.6         0.8          1.0         1.2 

Ratio  of  Q  and  V  to  their  Values  when  Pipe  is  Full 

FIG.  122. 


would  it  discharge  if  rilled  to  0.8  of  its  depth,  and  with  what 
velocity? 

Solution.  —  Fig.  122  shows  a  convenient  diagram  for  solving  a 
problem  of  this  kind  graphically.1  The  curve  marked  v  (velocity) 
is  plotted  from  Kutter's  simplified  formula 


100Vr\    ,- 
—  p  I  Vrs 
+  Vr/ 


for  a  value  of  b  of  0.35,  and  the  discharge  Q  from  the  formula 
Q  =  Av,  the  ordinates  to  the  curves  shown  in  the  figure  being  the 


Imhoff.  Taschenbuch  fur  Kanalisations  Ingenieure. 


140 


ELEMENTS  OF  HYDRAULICS 


ratio  of  the  depth  of  the  stream  to  the  diameter  of  the  pipe,  and 
the  abscissas  the  ratios  of  Q  and  v  respectively  to  their  values 
when  the  pipe  flows  full. 

To  apply  the  diagram  to  the  problem  under  consideration, 
observe  that  for  a  depth  of  0,8d  the  abscissa  of  the  discharge 
curve  is  unity,  and  consequently  the  discharge  for  this  depth  is 
the  same  as  when  the  pipe  is  completely  filled.  The  abscissa  of 
the  velocity  curve  corresponding  to  this  depth  0.8d  (i.e.,  with 
abscissa  0.8)  is  1.13,  and  consequently  the  velocity  at  this  depth 
is  1.13  X  9  =  10.17  ft.  per  second. 

Similar  diagrams  have  been  prepared  by  Imhoff  for  a  large 
variety  of  standard  cross  sections  and  are  supplemented  by 
other  .diagrams  or  charts  which  greatly  simplify  ordinary  sewer 
calculations. 

Problem  96. — In  the  Catskill  Aqueduct,  which  forms  part  of 
the  water  supply  system  of  the  City  of  New  York,  there  are  four 


FIG.  123. — Cut  and  cover  conduit,  Catskill  aqueduct. 

distinct  types  of  conduit;  the  cut-and-cover  type,  grade  tunnel, 
pressure  tunnel,  and  steel  pipe  siphon.  The  cut-and-cover  type, 
shown  in  section  in  Fig.  123,  is  55  miles  in  length,  and  is 
constructed  of  concrete  and  covered  with  an  earth  embankment. 
This  is  the  least  expensive  type,  and  is  used  wherever  the  eleva- 
tion and  nature  of  the  ground  permits. 

The  hydraulic  data  for  the  standard  type  in  open  cut  is  as 
follows : 


HYDROKINETICS 


141 


s  =  0.00021 

Depth  of 

Area  of 

Wetted 

Hydraulic 

„ 

flow 

flow 

perimeter  . 

radius 

in  feet 

in  sq.  ft. 

in  feet 

in  feet 

17.0 

241.0 

57.4 

4.20 

Full, 

16.2 

237.7 

50.8 

4.67 

Max.  cap. 

15.3 

230.9 

47.7 

4.84 

14.0 

217.6 

44.1 

4.92 

12.0 

192.6 

39.4 

4.88 

10.0 

163.2 

35;0 

4.65 

8.0 

130.7 

30.9 

4.24 

6.0 

97.1 

26.8 

3.61 

4.0 

62.2 

22.8 

2.72 

2.0 

27.0 

18.8 

1.47 

In  the  preliminary  calculations  the  relative  value  of  Chezy's 
coefficient  for  this  type  was  assumed  to  be  C  =  125.  Using 
this  value,  calculate  the  maximum  daily  discharge. 

Problem  97. — Where  hills  or  mountains  cross  the  line  of  the 
Aqueduct,  tunnels  are  driven  through  them  at  the  natural  eleva- 
tion of  the  Aqueduct  (Fig.  124).  There  are  24  of  these  grade 
tunnels,  aggregating  14  miles.  The  hydraulic  data  for  the  stand- 
ard type  of  grade  tunnel  is  as  follows : 


s  =  0.00037 

Depth  of 
flow 
in  feet 

Area  of 
flow 
in  sq.  ft. 

Wetted 
perimeter 
in  feet 

Hydraulic 
radius 
in  feet 

17.0 
16.25 
15.3 

.198.6 
195.6 

188.5 

52.2 
46.0 
42.7 

3.80 
4.25 
4.41 

Full, 
Max.  cap. 

14.0 

175.7 

39.3 

4.46 

12.0 

152.4 

35.0 

4.35 

10.0 

126.8 

31.0 

4.10 

8.0 

100.2 

26.9 

3.72 

6.0 

73.8 

22.9 

3.22 

4.0 

47.6 

18.9 

2.51 

- 

2.0 

21.0 

14.9 

1.49 

The  relative  value  of  Chezy's  coefficient  for  this  type  was  assumed 
in  the  preliminary  calculations  to  be  C  =  120.  Using  this  value, 
calculate  the  maximum  daily  discharge  and  the  corresponding 
velocity  of  flow. 

Problem  98. — Where  the  line  of  the  Aqueduct  crosses  broad 
and  deep  valleys  and  there  is  suitable  rock  beneath  them,  circular 


142 


ELEMENTS  OF  HYDRAULICS 


tunnels  are  driven  deep  in  the  rock  and  lined  with  concrete 
(Fig.  125).     There  are  seven  of  these  pressure  tunnels,  with  an 


FIG.  124. — Grade  tunnel,  Catskill          FIG.  125. — Pressure  tunnel,  Catskill 

aqueduct.  aqueduct. 

« 

aggregate  length  of  17  miles.     The  hydraulic   data  for  these 
pressure  tunnels  is  as  follows: 


Slope 

Diameter 

Area  of 
waterway 

Wetted 
perimeter 

Hydraulic 
radius 

0  .  00059 

14  ft.  6  in. 

165.1  sq.  ft. 

45.55  ft. 

3.625  ft. 

Assuming  the  relative  value  of  Chezy's  coefficient  to  be  C  =  120, 
calculate  the  velocity  of  flow  and  the  daily  discharge. 


\*Width  j.  5--*<    On  steep  hillside 


~-Hcif  sect/at  at  joint  -holes  shown 
dotted 

FIG.  126. — Steel  pipe  siphon,  Catskill  aqueduct. 

Problem  99. — In  valleys  where  the  rock  is  not  sound,  or  where 
for  other  reasons  pressure  tunnels  are  impracticable,  steel  pipe 
siphons  are  used  (Fig.  126).  These  are  made  of  steel  plates 
riveted  together,  from  7/16  to  3/4  of  an  inch  in  thickness,  and 


HYDROKINETICS  143 

are  9  ft.  and  11  ft.  in  diameter  respectively.  These  pipes  are 
embedded  in  concrete  and  covered  with  an  earth  embankment, 
and  are  lined  with  2  in.  of  cement  mortar  as  a  protection  to 
the  steel  and  also  for  the  sake  of  smoothness.  There  are  14  of 
these  siphons  aggregating  6  miles  in  length,  and  three  pipes  are 
required  for  the  full  capacity  of  the  Aqueduct.  Assuming  three 
mortar  lined  11-ft.  pipes,  having  a  relative  coefficient  of  0  =  120 
and  a  slope  s  =  0 . 00059,  calculate  the  velocity  of  flow  through 
them  and  the  maximum  daily  discharge. 

Problem  100. — A  broad  shallow  stream  has  naturally  a  depth 
of  3  ft.  and  a  longitudinal  slope  of  5  ft.  per  mile.  If  a  dam  8  ft. 
high  is  erected  across  the  stream,  determine  the  rise  in  level  one 
mile  up  stream  assuming  the  value  of  the  constant  C  in  Chezy's 
formula  as  75. 


SECTION  III 
HYDRODYNAMICS 

29.  PRESSURE  OF  JET  AGAINST  STATIONARY  DEFLECTING 

SURFACE 

Normal  Impact  on  Plane  Surface. — When  a  jet  of  water 
strikes  a  stationary  flat  plate  or  plane  surface  at  right  angles, 
the  water  spreads  out  equally  in  all  directions  and  flows  along 
this  plane  surface,  as  indicated  in  Fig.  127.  The  momentum 

of  the  water  after  striking  the 
surface  is  equal  to  the  sum  of  the 
momenta  of  its  separate  parti- 
cles, but  since  these  flow  off  in 
opposite  directions  their  alge- 
braic sum  is  zero.  Consequently 
the  entire  momentum  of  a  jet 
is  destroyed  by  normal  impact 
against  a  stationary  plane  sur- 
face. 

To  find  the  pressure,  P,  ex- 
erted by  the  jet  on  the  surface,  let  A  denote  the  cross  section 
of  the  jet  and  v  its  velocity.  Then  the  mass  of  water  flowing 
per  unit  of  time  is 

yAv 


FIG.  127. 


M  = 


(J 


and,  consequently,  from  the  principle  of  impulse  and  momentum, 

Pdt  =  Mv  =  iA^_ 


For  uniform  or  steady  flow,  P  is  constant,  and  if  M  denotes  the 
quantity  flowing  per  unit  of  time,  then  t  is  unity.  In  this 
case  the  above  expression  for  the  hydrodynamic  pressure  P 
of  the  jet  on  the  surface  becomes 


p  = 


(77) 


144 


HYDRODYNAMICS  145 

If  h  denotes  the  velocity  head,  then  h  =o~>   an<^   ^^   (^ 

may  be  written 

P  =  27  Ah.  (78) 

Relation  of  Static  to  Dynamic  Pressure. — If  the  orifice  is 
closed  by  a  cover  or  stopper,  then  the  hydrostatic  pressure 
P'  on  this  cover  is  approximately  equal  to  the  weight  of  a  column 
of  water  of  height  h  and  cross  section  A ;  and  consequently 

P'  =  7Ah.  (79) 

Comparing  Eq.  (78)  and  (79),  it  is  apparent  that  the  normal 
hydrodynamic  pressure  of  a  jet  on  an  external  plane  surface  is 
twice  as  great  as  the  hydrostatic  pressure  on  this  surface  would 
be  if  it  was  shoved  up  against  the  opening  so  as  to  entirely  close 
the  orifice. 

In  deriving  this  relation,  the  coefficient  of  efflux  is  assumed  to 
be  unity;  that  is,  the  area  A  of  the  jet  is  assumed  to  be  the 
same  as  that  of  the  orifice,  and  the  velocity  v  to  be  the  full  value 
corresponding  'to  the  head  h.  Since  the  coefficient  is  actually 
less  than  unity,  the  hydrodynamic  pressure  never  attains  the 
value  given  by  Eq.  (78).  For  instance,  in  the  case  of  flow 
from  a  standard  orifice,  if  A  denotes  the  area  of  the  orifice  and  a 
the  cross  section  of  the  jet,  then  from  Arts.  8  and  9 

a  =  0.62^1,  and  v  =  0.97 
Therefore  the  expression  for  P  becomes 

vav2  ,    — „, 

p  =  L    -  _  27(0.62^.)  (0.972A) 
9 

instead  of  2yAh,  as  given  by  Eq.  (78).  Note,  however,  that  this 
apparently  large  discrepancy  is  due  chiefly  to  the  fact  that  the 
area  A  in  Eq.  (78)  denotes  the  cross  section  of  the  jet,  whereas 
in  Eq.  (79)  it  denotes  the  area  of  the  orifice.  If  the  area  A  in 
both  expressions  denotes  the  cross  section  of  the  jet,  Eq.  (78) 
is  practically  true,  and  the  hydrodynamic  pressure  is  approxi- 
mately twice  the  hydrostatic  pressure  on  an  equal  area. 

Oblique  Impact  on  Plane  Surface. — If  a  jet  strikes  a  stationary 
plane  surface  obliquely,  at  an  angle  a  (Fig.  128),  the  axial  velocity 
v  of  the  jet  may  be  resolved  into  two  components,  v  sin  a  normal 
to  the  surface,  and  v  cos  a  tangential  to  the  surface.  If  the 
surface  is  perfectly  smooth,  the  water  flowing  along  the  surface 
experiences  no  resistance  to  motion,  and  the  pressure,  P,  exerted 
on  the  surface  is  that  corresponding  to  the  normal  velocity 
10 


146 


ELEMENTS  OF  HYDRAULICS 


component  v'  =  v  sin  a.     The  area  to  be  considered,  however, 
is  not  a  right  section,  A,  of  the  jet,  but  a  section  A'  normal  to 

the    component   v    sin   a,    as 

*A  '    /'/'W  *       m^icated  in  Fig.   128.      The 

total  pressure,  P,  exerted  on 
the  surface,  is  then 


or,  since  v'  =  v  sin  a  and  A'  = 
,  this  may  be  written  in 


sin 


the  form 


P  = 


g 


sin  a. 


(80) 


If  a  =  90°,  this  reduces  to  Eq.  (77). 

Axial  Impact  on  Surface  of  Revolution. — If  the  surface  on 
which  the  jet  impinges  is  a  surface  of  revolution,  coaxial  with  the 
jet  (Fig.  129),  then  in  this  case  also  the  particles  spread  out 


FIG.  129. 

equally  in  all  directions,  and  consequently  the  sum  of  the  momenta 
of  the  particles  in  the  direction  perpendicular  to  the  axis  of  the 
jet  is  zero.  The  velocity  of  any  particle  in  a  direction  parallel 
to  the  axis  of  the  jet,  however,  becomes  v  cos  a,  where  a  denotes 
the  angle  which  the  final  direction  taken  by  the  particles  makes 
with  their  initial  direction,  as  indicated  in  Fig.  129.  The  total 
initial  momentum  is  then 

Mv  =  ^f, 


HYDRODYNAMICS  147 

and  the  total  final  momentum  is 

yAv2 
Mv  cos  a  =  -     -  cos  a. 

g 

Therefore,  equating  the  impulse  to  the  change  in  momentum, 
we  obtain  the  relation 


p  = 


(i  _  COS  a). 


(81) 


For  a  jet  impinging  normally   on   a  plane  surface,  a  =  90°, 
and  this  expression  reduces  to  Eq.  (77). 

Complete  Reversal  of  Jet. — If  a  is  greater  than  90°,  then 
cos  a  becomes  negative  and  the  pressure  P  is  correspondingly 


FIG.  130. 

increased.  '  For  example,  if  the  direction  of  flow  is  completely 
reversed,  as  shown  in  Fig.  130,  then  a  =  180°,  cos  a  =  —  1, 
and  hence 

27  Av2 


P  = 


(82) 


The  hydrodynamic  pressure  in  this  case  is  therefore  twice  as 
great  as  the  normal  pressure  on  a  flat  surface,  and  four  times  as 
great  as  the  hydrostatic  pressure  on  a  cover  over  an  orifice  of 
the  same  area  as  the  cross  section  of  the  jet. 

Deflection  of  Jet. — When  a  jet  is  deflected  in  an  oblique 
direction,  the  final  velocity  v  may  be  resolved  into  components 
v  cos  a  and  v  sin  a,  as  indicated  in  Fig.  131.  The  component 
of  the  final  momentum  parallel  to  the  initial  direction  of  the  jet 
is  then 


Mv(l  —  cos  a)  = 


yAv' 


(1  —  COS  a), 


148  ELEMENTS  OF  HYDRAULICS 

and  the  horizontal  component,  H,  exerted  in  this  direction  is 


H  = 


g 


(l  —  COS  a). 


(83) 


Similarly,  the  component  of  the  final  momentum  perpendicular 
to  the  initial  direction  of  the  jet  is 

,,     .  jAv2    . 

Mv  sm  a  =  —  -  sin  a, 


V  cosef 


FIG.  131. 
and  the  vertical  component,  V,  exerted  in  this  direction  is 


V  = 


g 


, 
sin  a. 


(84) 


The  total  pressure  of  the  jet  on  the  deflecting  surface,  or  reaction 
of  the  surface  on  the  jet,  is,  then, 

P  =    V#2+F2   =  1^-  V(l  -  cos  a)2  +  sin2  a 

tJ 

which  simplifies  into 

P  =  — - —   V2(l    -  COS  a).  (85) 

o 

A  more  convenient  expression  for  P  may  be  obtained  by  using 
the  trigonometric  relation  +1 p^  -  =  sin  ^a,  by  means  of 


which  Eq.  (85)  may  be  written  in  the  form 


(86) 


HYDRODYNAMICS 


149 


Dynamic  Pressure  in  Pipe  Bends  and  Elbows. — When  a  bend 
or  elbow  occurs  in  a  pipe  through  which  water  is  flowing,  the 
change  in  directon  of  flow  produces  a  thrust  in  the  elbow,  as 
in  the  case  of  the  deflection  of  a  jet  by  a  curved  vane,  considered 
in  the  preceding  article.  From  Eq.  (85)  and  (86),  the  amount 
of  this  thrust  P  is 


g 


V2(l  -  cos  a)  = 


a 
sin  H, 


and  the  direction  of  the  thrust  evidently  bisects  the  angle  a, 
as  indicated  in  Fig.  132. 


FIG.  132. 


In  the  case  of  jointed  pipe  lines  if  the  angle  of  deflection  is 
large  or  the  velocity  of  flow  considerable,  this  thrust  may  be 
sufficient  to  disjoint  the  pipe  unless  provision  is  made  for  taking 
up  the  thrust  by  some  form  of  anchorage,  as,  for  example,  by 
filling  in  with  concrete  on  the  outside  of  the  elbow. 

30.  PRESSURE  EXERTED  BY  JET  ON  MOVING  VANE 

Relative  Velocity  of  Jet  and  Vane. — In  the  preceding  article 
it  was  assumed  that  the  surface  on  which  the  jet  impinged  was 
fixed  or  stationary.  The  results  obtained,  however,  remain 
valid  if  the  surface  moves  parallel  to  the  jet  in  the  same  or 
opposite  direction,  provided  the  velocity,  v,  refers  to  the 
relative  velocity  between  jet  and  surface.  Thus  if  the  surface 
moves  in  an  opposite  direction  to  the  jet  with  a  velocity  v', 
the  relative  velocity *of  jet  and  surface  is  v  +  v'  and  the  pressure 
is  correspondingly  increased,  whereas  if  they  move  in  the  same 
direction,  their  relative  velocity  is  v  —  v',  and  the  pressure  is 
diminished. 


150  ELEMENTS  OF  HYDRAULICS 

Work  Done  on  Moving  Vane. — Consider,  for  example,  the 
case  of  a  jet  striking  a  deflecting  surface  and  assume  first  that 
this  surface  moves  in  the  same  direction  as  the  jet  with  velocity 
vr  (Fig.  133).  Since  the  surface,  or  vane,  is  in  motion,  the 
mass  of  water,  M'}  reaching  the  vane  per  second  is  not  the  same 


FIG.  133. 

as  the  mass  of  water,  Mf  passing  a  given  cross  section  of  the  jet 
per  second.  That  is,  the  mass,  M,  issuing  from  the  jet  per 
second  is 


g 

whereas  the  mass,  M',  flowing  over  the  vane  per  second  is 
M,  =  yA(v  -^ 

g 

Therefore  the  components  of  the  force  acting  on  the  vane,  given 
by  Eq.  (83)  and  (84),  become  in  this  case 

H  =  M'(v  -  »')(!-  cos  a)  =  —  (v  -  v')\l  -  cos  a), 

yA        Q 

V  =  M'(v  -  */)  sin  a  =  J~(v  -  v'Y  sin  a. 

Since  the  motion  of  the  vane  is  assumed  to  be  in  the  direction 
of  the  component  H,  the  component  V,  perpendicular  to  this 
direction,  does  no  work.  The  total  work,  W,  done  on  the  vane 
by  the  jet  is  therefore 

W  =  Hv'  =  TAv/(v~V/)2(i  -  cos  a).  (88) 

»  ° 

Speed  at  which  Work  Becomes  a  Maximum.  —  The  condition 
that  the  work  done  shall  be  a  maximum  is 


^r  =  0  =        (1  -  cos  a)[(v  -  v'Y  -  2v'(v  -  v')], 
whence 

v'  =    -  (89) 


HYDRODYNAMICS  151 

% 

Substituting  this  value  of  vr  in  Eq.  (88),  the  maximum  amount 
of  work  that  can  be  realized  under  the  given  conditions  is  found 
to  be 

A- 

TT7  3     /  *V/1  *yAt;» 

Wmax  =  -  (v  —  gj   (1  —  cos  a)  =     27     (1  —cos  a). 

Maximum  Efficiency  for  Single  Vane.  —  The  efficiency  of  a, 
motor  or  machine  is  denned  in  general  as 

Useful  work 
Efflciency  =  Total  energy  available"  '  .(9o) 

Since,  in  the  present  case,  the  total  kinetic  energy  of  the  jet  is 


the  efficiency,  E,  becomes 

4jAv3 

-277  (     ~~ 
E  =  -  -  =        (l  _  cos  a). 

yAv3  27 


The  maximum  efficiency  occurs  when  a  —  180°,  in  which  case 
Emax  =  ~_  =  59-2  per  cent.  (91) 

Maximum  Efficiency  for  Continuous  Succession  of  Vanes.  — 

If  there  is  a  series  of  vanes  following  each  other  in  succession  so 
tthat  each  receives  only  a  portion  of  the  water,  allowing  this 
portion  to  expend  its  energy  completely  on  this  vane  before 
leaving  it,  then  the  mass  M'  in  Eq.  (87)  is  replaced  by  M,  and  the 
component  H  becomes 

H  ==  M(v  -  v')(l  -  cosa)=  yAv(v  ~^~  (I  -  cos  a). 
The  work  done  on  the  series  of  vanes  is  therefore 


W  .  HT'  =  .  (I  -  cos  a).  (92) 

***       .» 

The  condition  for  a  maximum  in  this  case  is 


dW  Nr/          /N         ., 

-^7-  =  0  =  l—  (1  -  cos  a)[(v  -  vf)  -  v'], 


whence 

v 


(93) 


152 


ELEMENTS  OF  HYDRAULICS 


Substituting  this  value  of  v'  inEq.  (92),  the  maximum  work  which 
can  be  realized  from  a  series  of  vanes  moving  parallel  to  the  jet  is 

yA  v2 
2 


w 

'' 


(•-I) 


/i  \  /-,  \ 

(1  —  COS  a)   =    —7 — (1— COS  a). 


Hence  the  efficiency  in  this  case  is 

! 

(1  —  COS  a) 


E 


2(7 


The  maximum  efficiency  therefore   occurs  when  a  —  180°,  its 
value  being 


x  =-    (2)  =  ioo   per  cent. 


(94) 


The  actual  efficiency  of  course  can  never  reach  this  upper  limit, 
as  the  conditions  assumed  are  ideal,  and  no  account  is  taken  of 
frictional  and  other  losses. 

Impulse  Wheel;  Direction  of  Vanes  at  Entrance  and  Exit. — 
In  general,  it  is  not  practicable  to  arrange  a  series  of  vanes  so 


Shaft 


FIG.  134. 

as  to  move  continuously  in  a  direction  parallel  to  the  jet.  As 
usually  constructed,  the  vanes  are  attached  to  the  circumference 
of  a  wheel  revolving  about  a  fixed  axis  (Fig.  134).  Let  co  denote 
the  angular  velocity  of  the  wheel  about  its  axis,  and  n,  r2  the  radii 


HYDRODYNAMICS  153 

of  the  inner  and  outer  edges  of  the  vanes.  Then  the  tangential 
or  linear  velocities  at  these  points,  say  HI  and  u^,  are 

u\  —  rico;  Uz  =  7*200. 

Now  let  Vi  denote  the  absolute  velocity  of  the  jet  at  entrance  to 
the  vane,  and  v%  its  absolute  velocity  at  exit.  Then  by  forming  a 
parallelogram  of  velocities  on  u\  and  Vi  as  sides,  the  relative 
velocity,  Wi,  between  jet  and  vane  at  entrance  is  determined, 
as  indicated  in  Fig.  134.  Similarly,  the  parallelogram  on  Uz,  Wz 
as  sides  determines  the  absolute  velocity,  v2,  at  exit.  In  order 
that  the  water  may  glide  on  the  vane  without  shock,  the  tip  of 
the  vane  at  entrance  must  coincide  in  direction  with  the 
vector  wi. 

Work  Absorbed  by  Impulse  Wheel. — Let  M  denote  the 
mass  of  water  passing  over  the  vane  per  second.  At  entrance 
the  velocity  of  this  mass  in  the  direction  of  motion  (i.e.,  its 
tangential  velocity)  is  v\  cos  a,  and  at  exit  is  Vz  cos  0,  where  a 
and  ]8  are  the  angles  indicated  in  Fig.  134.  The  linear  momentum 
of  the  mass  M  at  entrance  is  then  Mvi  cos  a,  and  its  angular 
momentum  is  MV&I  cos  a.  Similarly,  its  linear  momentum  at 
exit  is  Mvz  cos  0  and  its  angular  momentum  is  MV^TZ  cos  /3.  The 
total  change  in  angular  momentum  per  second,  that  is,  the 
amount  given  up  by  the  water  or  imparted  to  the  wheel,  is  then 

MviTi  cos  a  —  MvzTz  cos  |8. 

For  a  continuous  succession  of  vanes,  as  in  the  case  of  an 
ordinary  impulse  wheel,  the  mass  M  is  the  total  amount  of  water 
supplied  by  the  jet  per  second.  Hence,  if  T  denotes  the  total 
torque  exerted  on  the  wheel,  by  the  principle  of  angular  impulse 
and  momentum,  remembering  that  M  is  the  mass  of  water  flowing 
per  unit  of  time,  and  consequently  that  the  time  is  unity, 

T  =  M(viii  cos  a-v2r2  cos  0).  (95) 

The  total  work  imparted  to  the  wheel  is 

W=  Tu, 

or,  since  M  =  -    — >   u\  —  riw,  u2  =  r2co,  the  expression  for  the 

y 
work  becomes 

W  =  Tco  = 1  (UiVi  COS  a  -  U2V2  COS  |8).  (96) 

These  relations  will  be  applied  in  Art.  34  to  calculating  the 
power  and  efficiency  of  certain  types  of  impulse  wheels. 


154 


ELEMENTS  OF  HYDRAULICS 


31.  REACTION  OF  A  JET 

Effect  of  Issuing  Jet  on  Equilibrium  of  Tank. — Consider  a  closed 
tank  containing  water  or  other  liquid,  and  having  an  orifice 
in  one  side  closed  by  a  cover.  When  the  cover  is  removed  the 
equilibrium  of  water  and  tank  will  be  destroyed.  At  the  instant 
of  removal  this  is  due  to  the  disappearance  of  the  pressure  pre- 
viously exerted  on  the  cover  considered  as  part  of  the  tank. 
After  the  jet  has  formed  and  a  steady  flow  has  been  set  up,  as- 
suming that  the  depth  of  water  is  maintained  constant  by  supply- 
ing an  amount  equal  to  that  flowing  out,  as  indicated  in  Fig.  135, 
the  pressure  within  the  fluid  and  on  the  walls  of  the  tank  will  not 
regain  its  original  static  value,  since,  in  accordance  with  Ber- 
noulli's theorem,  an  increase  in  velocity  must  be  accompanied  by 
a  corresponding  decrease  in  pressure. 

Energy  of  Flow  Absorbed  by  Work  on  Tank. — To  calculate  the 
effect  of  the  flow  on  the  equilibrium  of  the  tank,  suppose  that  the 


FIG.  135. 

tank  is  moved  in  the  direction  opposite  to  that  of  the  jet  and  with 
the  same  velocity,  v,  as  that  of  the  jet.  Then  the  relative  velocity 
of  the  jet  with  respect  to  the  tank  is  still  v,  but  the  absolute 
velocity  of  the  jet  is  zero  and  consequently  its  kinetic  energy  is 
also  zero.  If  h  denotes  the  head  of  water  on  the  orifice  (Fig.  135) 
and  Q  the  quantity  of  water  flowing  per  second,  then  its  loss  in 
potential  energy  per  second  is  yQh.  Moreover,  while  this 
volume  of  water  Q  moved  with  the  tank,  it  had  a  velocity  v, 

and  therefore  possessed  kinetic  energy  of  amount  -~ —      The 

total  energy  given  up  by  the  water  in  flowing  from  the  tank  is 
then 


HYDRODYNAMICS  155 

v2 
or,  since  h  =  ~—  approximately,  these  terms  are  equal,  and  the 

ZQ 
total  energy  lost  by  the  water  becomes 

yQv2 
Energy  given  up  =  - 

i/ 

Now  let  P  denote  the  reaction  of  the  jet,  that  is,  the  resultant  of 
all  the  pressure  exerted  on  the  tank  by  the  water  except  that  due 
to  its  weight.  Then,  since  the  distance  traversed  by  the  force  P 
in  a  unit  of  time  is  the  velocity  v  of  the  tank,  by  equating  the 
work  done  by  P  to  the  energy  given  up  by  the  water,  we  have 


whence 


=  =  = 

g         g 


The  reaction  P  is  therefore  twice  the  hydrostatic  pressure  due  to 
the  head  h. 

This  is  also  apparent  from  the  results  of  Art.  27,  since  the  pres- 
sure of  a  jet  on  a  fixed  surface  close  to  the  orifice  must  be  equal  to 
its  reaction  on  the  vessel  from  which  the  jet  issues.  The  actual 
reaction  of  the  jet  is  of  course  somewhat  less  than  its  theoretical 
value,  as  given  by  the  relation  P  =  2yAh,  since  there  are  various 
losses,  due  to  internal  friction,  etc. 

Principle  of  Reaction  Turbine.  —  In  order  for  the  tank  to  retain 
its  uniform  velocity,  v,  a  resistance  of  amount  P  must  constantly 
be  overcome,  for  if  the  resistance  is  less  than  this  amount  the 
motion  will  be  accelerated.  It  is  apparent,  therefore,  that  by  a 
proper  choice  of  the  velocity,  v,  of  the  tank  it  is  possible  to  utilize 
almost  the  entire  energy  of  the  jet  in  overcoming  a  resistance 
coupled  up  with  the  tank.  This  is  the  principle  on  which  the 
reaction  turbine  is  based,  as  explained  in  Art.  35. 

It  should  be  noted  that  if  the  water  flowing  out  is  continually 
replaced  from  above,  half  of  the  available  energy  must  be  used  in 
giving  the  supply  water  the  same  velocity  as  the  tank.  The  use- 
ful work  is  therefore  reduced  to  one-half  the  previous  amount, 
and  the  available  energy  is  only  that  due  to  the  velocity  head  h. 

Barker's  Mill.  —  The  simplest  practical  application  of  the 
reaction  of  a  jet  is  the  apparatus  known  as  Barker's  mill  (Fig.  136). 
In  this  apparatus  water  flows  from  a  tank  into  a  hollow  vertical 
arm,  or  spindle,  pivoted  at  the  lower  end,  and  from  this  into  a 


156 


ELEMENTS  OF  HYDRAULICS 


SIDE  ELEVATION 


o 


horizontal  tubular  arm,  having  two  orifices  near  the  ends  on 
opposite  sides.  The  jets  issue  from  these  orifices,  and  their 
reactions  cause  the  horizontal  arm  to  rotate,  driving  the  central 

spindle  from  which  the  power  is  taken 
off  by  a  belt  and  pulley. 

The  steam  turbine  invented  by 
Hero  of  Alexandria  in  the  first  cen- 
tury B.  C.,  is  an  almost  identical 
arrangement,  the  motive  power  in 
this  case  being  due  to  the  reaction  of 
a  jet  of  steam  instead  of  a  jet  of 
water. 

32.  TYPES  OF  HYDRAULIC  MOTORS 

Current  Wheels. — There  are  three 
general  types  of  hydraulic  motors, 
namely 

1.  Current  and  gravity  wheels, 

2.  Impulse  wheels  and  turbines, 

3.  Reaction  turbines. 

The  current  wheel  is  the  oldest  type 
of  prime  mover,  and  in  its  primitive 
form  consisted  of  a  large  vertical 
wheel,  with  a  set  of  paddles  or 

buckets  attached  to  its  circumference,  and  so  placed  in  a 
running  stream  that  the  current  acting  on  the  lower,  or  im- 
mersed, portion  produced  revolution  of  the  wheel.  A  later  im- 
provement consisted  in  placing  the  wheel  at  the  foot  of  a  water 
fall  and  conducting  the  water  by  a  flume  to  the  top  of  the  wheel, 
the  action  of  the  water  in  this  case  being  due  almost  entirely  to  its 
weight  (Art.  33). 

Impulse  Wheels. — The  impulse  wheel,  as  its  name  indicates, 
is  designed  to  utilize  the  impulsive  force  exerted  by  a  jet  moving 
with  a  high  velocity  and  striking  the  wheel  tangentially.  The 
wheel,  or  runner,  in  this  case  carries  a  series  of  curved  buckets  or 
vanes  which  discharge  into  the  atmosphere.  A  feature  of  this 
type  is  that  the  runner  rotates  at  a  high  velocity  and  can  there- 
fore be  made  of  comparatively  small  diameter.  The  two  prin- 
cipal types  of  impulse  wheel  are  the  Girard  impulse  turbine, 
which  originated  in  Europe,  and  the  Pelton  wheel,  which  was 
developed  in  the  United  States  (Art.  34). 


PLAN 


F!G.    136. 


HYDRODYNAMICS 


157 


Reaction  Turbines. — The  reaction  turbine  depends  chiefly  on 
the  reaction  exerted  by  a  jet  on  the  vessel  from  which  it  flows, 
which  in  this  case  is  the  passage  between  the  vanes  on  the 
runner.  In  an  impulse  wheel  the  energy  of  the  water  as  it  enters 
the  wheel  is  entirely  kinetic,  and  as  there  is  free  circulation  of  air 
between  the  vanes  and  they  discharge  into  the  atmosphere,  the 
velocity  of  the  water  is  that  due  to  the  actual  head.  In  a 
reaction  turbine  the  energy  of  the  water  as  it  enters  the  wheel  is 
partly  kinetic  and  partly  pressure  energy,  and  as  the  water 
completely  fills  the  passages  between  the  vanes,  its  velocity  at 
entrance  may  be  either  greater  or  less  than  that  due  to  the  static 
head  at  that  point.  A  feature  of  the  reaction  turbine  is  that 
it  will  operate  when  completely  submerged. 

Classification  of  Reaction  Turbines. — Reaction  turbines  are 
subdivided  into  four  classes,  according  to  the  direction  in  which 
the  water  flows  through  the  wheel.  These  are 

(1)  Radial  outward  flow  turbines, 

(2)  Radial  inward  flow  turbines, 

(3)  Parallel  or  axial  flow  turbines, 

(4)  Mixed  flow  turbines,  the  direction  of  flow  being  partly 
radial  and  partly  axial,  changing  from  one  to  the  other 
in  passing  over  the  vanes  (Art.  35). 

Classification  of  Hydraulic  Motors. — The  following  tabulated 
classification  is  useful  as  a  basis  for  the  description  of  the  various 
types  of  hydraulic  motors  given  in  Arts.  33,  34  and  35. 


Current  and  Gravity  Wheels: 

Utilizes  impact  of  current  or  weight 
of  the  water. 

Impulse  Wheels  and  Turbines: 
Utilizes  kinetic  energy  of  jet  at 
high  velocity.  Suitable  for  limited 
amount  of  water  under  high  head. 
Ordinarily  used  for  heads  from  300 
ft.  to  3000  ft. 

Reaction  Turbines: 

Utilizes  both  kinetic  and  pressure  en- 
ergy of  water.  Suitable  for  large 
quantities  of  water  under  low  or  me- 
dium head.  Ordinarily  used  for 
heads  from  5  to  500  ft. 


Current  wheel, 
Undershot  wheel  (Poncelet), 
Breast  wheel, 
Overshot  wheel. 


Girard    turbine    (European), 
Pelton  wheel  (American). 


Radial  inward  flow  (Francis  type), 
Radial  outward  flow  (Fourneyron 

type), 

Parallel  or  axial  flow  (Jonval  type), 
Mixed  flow  (American  type). 


158  ELEMENTS  OF  HYDRAULICS 

33.  CURRENT  AND  GRAVITY  WHEELS 

Current  Wheels. — The  vertical  current  wheel,  mentioned  in 
Art.  32,  was  the  earliest  type  of  hydraulic  motor,  dating  from 
prehistoric  times,  although  they  are  still  in  use  in  China  and 
Syria. 

Undershot  Wheels. — The  first  improvement  consisted  in 
confining  the  water  in  a  sluice  and  delivering  it  directly  on  the 
vanes.  This  type  was  known  as  the  Undershot  wheel,  and  was  in 
common  use  until  about  the  year  1800  A.  D.  (Fig.  137).  Flat 
radial  vanes  were  used  with  this  type,  for  which  the  maximum 
theoretical  efficiency  was  50  per  cent.,  the  velocity  of  the  vanes  to 
realize  this  efficiency  being  one-half  the  velocity  of  the  stream,  as 
explained  in  Art.  29.  The  actual  efficiency  of  such  wheels  was 
much  lower,  being  only  from  20  to  30  per  cent. 


^^^^^^^^^^^^^^%^^% 

FIG.  137.  FIG.  138. 

Poncelet  Wheels. — Undershot  wheels  were  greatly  improved 
by  Poncelet  who  curved  the  vanes,  so  that  the  water  entered 
without  shock  and  was  discharged  in  a  nearly  vertical  direction 
(Fig.  138).  The  water  thus  exerted  an  impulse  on  the  vanes 
during  the  entire  time  it  remained  in  the  wheel,  thereby  raising 
the  actual  efficiency  to  about  60  per  cent.  Undershot  wheels  of 
the  Poncelet  •  type  are  adapted  to  low  falls,  not  exceeding  7  ft. 
in  height. 

Breast  Wheels. — A  modification  of  the  undershot  wheel 
is  the  Breast  wheel,  the  water  being  delivered  higher  up  than  for 
an  ordinary  undershot  wheel,  and  retained  in  the  buckets  during 
the  descent  by  means  of  a  breast,  or  casing,  which  fits  the  wheel  as 
closely  as  practicable  (Fig.  139).  Wheels  of  this  type  are  known 
as  high  breast,  breast,  and  low  breast  according  as  the  water  is 
delivered  to  the  wheel  above,  at,  or  below  the  level  of  the  center 


HYDRODYNAMICS  159 

of  the  wheel.  The  high  breast  wheel  operates  almost  entirely  by 
gravity,  that  is,  by  the  unbalanced  weight  of  the  water  in  the 
buckets,  its  efficiency  being  from  70  to  80  per  cent.  Breast  and 
low  breast  wheels  operate  partly  by  gravity  and  partly  by  impulse, 
the  efficiency  varying  from  about  50  per  cent,  for  small  wheels  to 
80  per  cent,  for  large  wheels.  This  type  was  in  use  until  about 
1850. 

Overshot  Wheels. — A  more  recent  type  is  the  Overshot  wheel, 
the  characteristic  of  this  type  being  that  the  water  is  delivered  at 
the  top  of  the  wheel  by  a  sluice,  as  indicated  in  Fig.  140.  For 


I  Head  Race, 

Head 


Breast 


FIG.  139.  FIG.  140. 

maximum  efficiency  the  diameter  of  the  wheel  should  be  nearly 
equal  to  the  height  of  the  fall,  the  efficiency  for  well-designed 
overshot  wheels  ranging  from  70  to  85  per  cent.,  which  is  nearly 
as  high  as  for  a  modern  turbine.  An  overshot  wheel  at  Troy, 
N.  Y.,  is  62  ft.  in  diameter,  22  ft.  wide,  weighs  230  tons,  and 
develops  550  H.P.  Another  on  the  Isle  of  Man  is  72  ft.  in 
diameter  and  develops  150  H.P. 

34.  IMPULSE  WHEELS  AND  TURBINES 

Pelton  Wheel. — The  intermediate  link  between  the  old  type 
of  water  wheel  and  the  modern  impulse  wheel  was  the  Hurdy 
Gurdy,  which  was  introduced  into  the  mining  districts  of  Cali- 
fornia about  1865.  This  somewhat  resembled  the  old  current 
wheel,  being  vertical  with  flat  radial  vanes,  but  differed  from  it 
in  that  it  was  operated  by  a  jet  impinging  on  the  vanes  at  high 
velocity.  The  maximum  theoretical  efficiency  of  the  Hurdy 
Gurdy  was  50  per  cent.  (Art.  29),  while  its  actual  efficiency  varied 
from  25  to  35  per  cent. 


160 


ELEMENTS  OF  HYDRAULICS 


The  substitution  of  curved  buckets  for  the  flat  radial  vanes 
was  the  great  improvement  which  converted  the  Hurdy  Gurdy 
into  the  Pelton  wheel.  The  construction  of  the  bucket  is  shown 
in  Fig.  141,  the  jet  being  divided  by  the  central  ridge  and  each 
half  deflected  through  nearly  180°.  Evidently  the  angle  of 
deflection  must  be  slightly  less  than  180°,  so  that  the  discharge 
from  one  bucket  may  clear  the  one  following.  A  later  improve- 
ment is  the  Doble  bucket,  also  shown  in  Fig.  141,  each  half  of 


Pelton  Bucket 


Doble  Bucket 
FIG.  141. 

which  is  ellipsoidal  in  form,  with  part  of  the  outer  lip  cut  away 
so  as  to  clear  the  jet  when  coming  into  action. 

The  relation  of  the  jet  to  the  wheel  is  shown  in  Fig.  142,  the 
type  there  shown  being  arranged  with  a  deflecting  nozzle  for  eco- 
nomic regulation.  A  more  recent  type  of  Pelton  wheel  is  shown  in 
Fig.  143,  the  features  of  this  type  being  the  Doble  buckets  and  the 
so-called  chain  type  of  attachment  of  the  buckets. 

One  of  the  most  important  features  of  construction  in  this  type 
of  impulse  wheel  is  the  needle  valve  for  regulating  the  flow.  The 


HYDRODYNAMICS 


161 


T 


FIG.  142. 


FIG.  143.— Pelton-Doble  runner,  diameter  153  in.,  weight  28,000  Ibs. 


162  ELEMENTS  OF  HYDRAULICS 

cross  section  shown  in  Fig.  144  indicates  the  location  of  the  needle 
valve  with  respect  to  the  nozzle.  The  methods  of  operating 
the  valve  and  of  elevating  and  depressing  the  nozzle  are  shown  in 
Fig.  145.  This  form  of  nozzle  under  the  high  heads  ordinarily 
used  gives  a  very  smooth  and  compact  jet,  as  shown  by  the 
instantaneous  photograph  reproduced  in  Fig.  146. 

Efficiency  of  Pelton  Wheel. — If  the  jet  was  completely  reversed 
in  direction  and  the  speed  of  the  buckets  was  one-half  that  of 
the  jet,  the  theoretical  efficiency  of  the  Pelton  wheel  would  be 
unity,  or  100  per  cent.,  as  shown  by  Art.  30,  Eq.  (94).  This  is 


FIG.  144. 

also  apparent  from  other  considerations;  for  if  the  velocity  of 
the  jet  is  v  and  that  of  the  buckets  is  -~  ,  then  the  velocity  of  the 

water  relative  to  the  lowest  bucket  is  v  —  ~-,    or   -~    (Fig.    147). 

Zi  Z 

Therefore  at  exit  the  water  is  moving  with  velocity  -^  relative 
to  the  bucket  while  the  bucket  itself  is  moving  in  the  opposite 
direction  with  velocity  -~  *  Hence  the  absolute  velocity  of  the 

water  at  exit  is  -^    —  -H  ,  or  zero,  and  therefore,  since  the  total 

kinetic  energy  of  the  water  has  been  utilized,  the  theoretical 
efficiency  of  the  wheel  is  unity.  As  a  matter  of  fact  there  are 
hydraulic  friction  losses  to  be  taken  into  account  and  also  the 
direction  of  flow  is  not  completely  reversed.  The  efficiency  of 
the  Pelton  wheel  has  been  found  in  a  number  of  authentic  tests 
to  exceed  86  per  cent.  The  actual  efficiency  in  operation  de- 


HYDRODYNAMICS 


163 


^L 


FIG.  145. — Pelton  regulating  needle  nozzle. 


FIG.  146. 


164 


ELEMENTS  OF  HYDRAULICS 


pends  of  course  on  the  particular  hydraulic  conditions  under 
which  the  wheel  operates.  A  good  idea  of  what  may  be  ex- 
pected in  practice,  however,  is  given  by  the  following  data,  ob- 
tained from  a  unit  of  4000  KW  normal  capacity,  operating 
under  a  head  of  1300  feet: 


Load  in  KW 

Percentage  of  normal 
capacity 

Wheel  efficiency 

5000 
4000 
3000 
2000 

125% 
100% 

75% 
50% 

81% 
83% 

82% 

70% 

FIG.  147. 

Since  the  Pelton  wheel  operates  by  utilizing  the  kinetic  energy 
of  the  water,  it  is  best  adapted  to  a  small  discharge  under  a  high 
head. 

Girard  Impulse  Turbine. — A  somewhat  different  type  of 
impulse  wheel  has  been  developed  in  Europe,  known  as  the 
Girard  impulse  turbine.  In  this  type  the  shaft  may  be  mounted 
either  vertically  or  horizontally,  and  the  flow  may  be  either  radial 
or  axial.  The  type  shown  in  Fig.  148  is  arranged  for  radial 
flow,  with  vertical  shaft.  The  construction  is  practically  the 
same  in  all  cases,  the  water  entering  through  a  pipe  B,  as  shown 
in  Fig.  148,  and  proceeding  through  one  or  more  guide  passages 
C,  which  direct  the  water  onto  the  vanes.  The  quantity  of 
water  admitted  to  the  vanes  is  regulated  by  some  kind  of  gate, 
that  indicated  in  Fig.  148  being  a  sliding  gate  operated  by  a  rack 
and  pinion  not  shown  in  the  figure. 


HYDRODYNAMICS  165 

As  the  vanes  are  more  oblique  at  exit  than  at  entrance,  they 
are  necessarily  closer  together  at  exit.  To  prevent  choking, 
it  is  therefore  necessary  to  widen  the  vanes  laterally  at  exit,  as 
shown  in  elevation  in  Fig.  148.  As  the  water  discharges  under 
atmospheric  pressure,  ventilating  holes  are  made  through  the 
sides  of  the  vanes  at  the  back  to  allow  free  admission  of  air. 


FIG.  148. 

Power  and  Efficiency  of  Girard  Turbine. — The  power  and 
efficiency  of  a  Girard  turbine  may  easily  be  calculated  from  the 
results  of  Art.  30.  Using  the  same  notation,  as  indicated  on 
Fig.  134,  from  Eq.  (96),  Art.  30,  the  work  done  per  second  on  the 
wheel  is  given  by  the  relation 

Work  per  second  =  -       -  (u\  Vi  cos  a  —  u2Vz  cos  £)* 


166 


ELEMENTS  OF  HYDRAULICS 


Since  the  water  is  under  atmospheric  pressure,  the  absolute 
velocity  Vi  of  the  water  at  entrance  is  calculated  from  the 
effective  head  h  by  means  of  the  relation  vi  =  ^J2gh.  From 
Fig.  149  we  have  by  geometry 

Vi  cos  a  =  Ui  +  Wi  cos  6, 
v%  cos  /3    =  u%  —  Wz  cos  tf>, 

and  from  the  law  of  cosines 

v^  =  u^  +  Wi2  +  2u iWi  cos  6. 

COS   </>. 


FIG.  149. 


Now  the  total  energy  imparted  to  the  wheel  per  second  is  the 
difference  in  kinetic  energy  at  entrance  and  exit,  namely 


Therefore,  equating  this  to  the  expression  for  the  work  done  per 
second,  as  given  above,  we  have 


COS    a   — 


COS 


Substituting    in    this    equation    the    values    given    above    for 
Vi  cos  a,  Vz  cos  j3,  Vi2  and  v22  and  reducing,  the  result  is  finally 


.2    _ 


(97) 


It  is  evident  that  the  efficiency  will  be  greater  the  more  nearly 
the  jet  is  reversed  in  direction,  that  is,  the  smaller  <p  becomes, 
or  what  amounts  to  the  same  thing,  the  smaller  the  absolute 


HYDRODYNAMICS  167 

velocity  v2  of  the  water  at  exit.  However,  <p  cannot  be  decreased 
indefinitely  as  it  is  necessary  to  provide  a  sufficient  area  at  exit 
to  carry  the  discharge.  For  a  given  value  of  <p,  v2  will  attain 
nearly  its  minimum  value  when  u2  =  w2.  In  this  case,  however, 
by  Eq.  (97)  we  have  HI  =  Wi,  in  consequence  of  which 

0  =  2a,  and  |  =  90°  -  0, 

and  hence 

<p  ?>. 

v2  =  2u2  sin  TT,  and  Ui  = 


cos  a 


Now  the  peripheral  velocity  of  the  inner  and  outer  ends  of  the 
vanes  in  terms  of  the  angular  velocity  co  of  the  runner  is  given  by 
the  relations  u\  =  TI  «,  u2  =  r2co,  whence 


u2        r2 
-  =  — '     or  u2  = 


Substituting  this  value  of  u2  in  the  expression  given  above  for 
02,  we  have 

2uir2  sin  ^       Vir2  sin  ^ 


.    <p 
v2  =  2u2  sin  ^-  = 


2  ri  r\  cos  a 

Consequently  if  Q  denotes  the  quantity  of  water  discharged  per 
second,  the  energy  utilized  per  second  is 

Energy  per  second  =  ^  (Vl*  -  vft  =  ^ 

Substituting  y  =  62.4  and  dividing  by  550,  the  expression  for  the 
horse  power  of  the  wheel  is  therefore 


Since  the  total  kinetic  energy  available  is    ^l  >  the  efficiency, 
E,  of  the  wheel,  as  defined  by  Eq.  (90),  Art.  30,  is 


a 


168  ELEMENTS  OF  HYDRAULICS 

Since  E  is  less  than  unity,  it  is  evident  that  the  maximum 
theoretical  efficiency  is  always  less  than  100  per  cent.,  and  also 
that  the  efficiency  is  greater  the  smaller  the  angles  a  and  <p. 

In  practice  the  angle  a  is  usually  between  20°  and  25°;  ^be- 
tween 15°  and  20°;  and  the  ratio  --  between  1.15  and  1.25. 

Assuming  as  average  numerical  values 

a   =  20°,    <p  =  15°,     -  =  1.15, 

and  substituting  these  values  in  Eq.  (99),  the  theoretical  efficiency 
of  the  wheel  in  this  case  is  found  to  be 

[sin  7  5°H 2 
1.15  —       '  o      =  97.5  per  cent,  approximately. 
COS  ZU  J  * 

This  efficiency  is,  of  course,  merely  ideal  as  it  takes  no  account 
of  hydraulic  friction  losses. 

The  Girard  type  of  impulse  turbine  was  formerly  manufac- 
tured in  this  country  by  the  Stilwell-Bierce  and  Smith- Vaile 
Co.  (now  the  Platt  Iron  Works  Co.,  of  Dayton,  Ohio)  under  the 
trade  name  of  the  "Victor  High  Pressure  Turbine. "  In  a  test 
of  a  45  inch-wheel  of  this  type  installed  in  the  power  plant  of 
the  Quebec  Railway  Light  and  Power  Co.,  Montmorency  Falls, 
Quebec,  with  a  rated  capacity  of  1000  horse  power  under  a  head 
of  195  feet  at  a  speed  of  286  r.p.m.,  a  maximum  efficiency  of 
78.38  per  cent,  was  attained. 

In  another  test  of  a  25-inch  Victor  wheel  installed  in  the 
Napanoch  Power  Station  of  the  Honk  Falls  Power  Co.,  Ellens- 
ville,  N.  Y.,  with  a  rated  capacity  of  500  horse  power  under  145 
feet  head  at  a  speed  of  480  r.p.m.,  a  maximum  efficiency  of  84.2 
per  cent,  was  attained. 

The  average  efficiency  of  Victor  wheels  in  plants  installed  is 
said  by  the  manufacturers  to  vary  from  70  to  80  per  cent.,  de- 
pending on  the  design  of  unit. 

In  this  type  of  unit,  no  draft  tube  is -used  and  consequently 
that  portion  of  the  head  from  the  center  line  of  the  wheel  to 
the  level  of  the  tail  race  is  lost.  Various  attempts  have  been 
made,  both  with  this  arid  other  types  of  impulse  wheel,  to  re- 
gain at  least  part  of  this  lost  head  by  means  of  an  automatic- 
ally regulated  draft  tube,  designed  to  keep  the  water  at  a  cer- 
tain fixed  level  beneath  the  runner,  but  this  feature  has  never 
proved  successful  in  operation. 


HYDRODYNAMICS 
36.  REACTION  TURBINES 


169 


Historical  Development. — In  Art.  29  it  was  shown  that  a  jet 
flowing  from  a  vessel  or  tank  exerted  a  pressure  or  reaction  on  the 
vessel  from  which  it  flows.  A  simple  application  of  this  principle 
was  shown  in  Barker's  mill,  Fig.  136,  in  which  the  reactions  of 


Radial 
Outward  Flow 


Fourneyron 
Type 


FIG.  150. 

two  jets  caused  a  horizontal  arm  to  revolve.  Later  this  device 
was  improved  by  curving  the  arms  so  that  the  jets  issued  directly 
from  the  ends  of  the  arms  instead  of  from  orifices  in  the  side,  and 
in  this  form  it  was  known  as  the  Scotch  mill.  Subsequently  the 


Axial.  Flow 


FIG.  151. 

number  of  arms  was  increased  and  the  openings  enlarged,  until 
it  finally  developed  into  a  complete  wheel. 

In  1826  a  French  engineer,  Fourneyron,  placed  stationary  guide 
vanes  in  the  center  to  direct  the  water  onto  the  runner,  or  wheel, 
the  result  being  the  first  reaction  turbine,  now  known  as  the 


170 


ELEMENTS  OF  HYDRAULICS 


Fourneyron  or  outward  flow  type  (Fig.  150).  This  type  was 
introduced  into  the  United  States  in  1843. 

A  later  modification  of  design  resulted  in  the  axial  or  parallel 
flow  turbine,  known  as  the  Jonval  type,  which  was  also  of  European 
origin,  and  was  introduced  into  the  United  States  about  1850 
(Fig.  151). 

A  crude  form  of  inward  flow  turbine  was  built  in  the  United 
States  as  early  as  1838.  Subsequently  the  design  was  greatly 
improved  by  the  noted  American  hydraulic  engineer,  J.  B.  Francis, 
and  it  has  since  been  known  as  the  Francis  type  (Fig.  152). 


FIG.  152. 


Fig.  153  shows  a  runner  of  the  Francis  type  used  in  the  plant 
of  the  Ontario  Power  Co.  at  Niagara  Falls.  These  are  double 
central  discharge,  or  balanced  twin  turbines,  designed  to  deliver 
13,400  H.P.  per  unit,  under  180  ft.  head.  The  runners  are  of 
bronze,  82-3/4  in.  in  diameter;  the  shafts  24  in.  maximum 
diameter;  and  the  housings  of  reinforced  steel  plate  16  ft.  in 
diameter,  spiral  in  elevation  and  rectangular  in  plan,  as  shown  in 
sectional  detail  in  Fig.  154.  A  cross  section  of  the  power  house 
in  which  these  turbines  are  installed  is  shown  in  Fig.  155. 

Mixed  Flow,  or  American,  Type. — The  mixed  flow  turbine, 
or  American  type,  is  a  modification  of  the  Francis  turbine  re- 
sulting from  a  demand  for  higher  speed  and  power  under  low 
heads.  Higher  speed  could  only  be  obtained  by  using  runners  of 
smaller  diameter,  which  meant  less  power  if  the  design  was  un- 
altered in  other  respects.  To  increase  the  capacity  of  a  runner  of 
given  diameter  the  width  of  the  runner  was  increased,  fewer  vanes 
were  used,  and  they  were  extended  further  toward  the  center. 
As  this  decreased  the  discharge  area,  the  vanes  were  curved  so  as 
to  discharge  the  water  axially  (Fig.  156).  In  a  standard  turbine 


HYDRODYNAMICS 


171 


FIG.  153. 


FIG.  154. 


172 


ELEMENTS  OF  HYDRAULICS 


HYDRODYNAMICS  173 

of  this  type,  the  water  from  the  conduit  or  penstock,  after  pass- 
ing through  the  shut-off  valves,  enters  a  cast-iron  or  cast-steel  cas- 
ing of  spiral  form  encircling  the  runner,  by  which  it  is  delivered 
to  the  whole  circumference  of  the  runner  at  a  uniform  velocity 
(Fig.  157).  The  detail  of  the  gate  work  for  regulating  the 
admission  of  water  to  the  runner  is  shown  in  Fig.  158,  and  the 
entire  turbine  unit  is  shown  in  perspective  in  Fig.  159. 


FIG.  156. — American  turbine  runners  built  by  the  Allis  Chalmers 
Manufacturing  Co. 


Use  of  Draft  Tube. — In  a  reaction,  or  pressure,  turbine  the 
passages  between  the  vanes  are  completely  filled  with  water, 
and  since  this  is  the  case,  it  will  run  submerged.  By  the  use  of 
a  draft  tube  or  suction  tube,  invented  by  Jonval  in  1843,  it  is 
possible,  however,  to  set  the  turbine  above  the  level  of  the  tail 
water  without  losing  head  (Fig.  160).  This  is  due  to  the  fact 
that  the  pressure  at  the  upper  end  of  the  draft  tube  is  enough 
less  than  atmospheric  to  compensate  for  the  loss  of  hydrostatic 
pressure  at  the  point  of  entrance  to  the  wheel.  In  fact,  by  the 
use  of  a  flaring  draft  tube  the  efficiency  may  be  slightly  increased. 
The  chief  advantage  of  a  draft  tube,  however,  is  that  its  use  per- 
mits of  setting  the  turbine  in  a  more  accessible  position  without 
any  sacrifice  of  head. 

The  effect  of  using  a  draft  tube  may  be  explained  mathe- 
matically as  follows: — In  Figure  161  let  B  refer  to  the  point  at 
which  water  enters  the  turbine,  and  C  to  its  point  of  exit  into 


174 


ELEMENTS  OF  HYDRAULICS 


the  draft  tube.     Then  writing  out  Bernoulli's  theorem  between 
the  points  A  and  B,  we  obtain  the  relation 


PA 


29 


Since  the  intake  tube  is  assumed  to  be  of  uniform  cross  section, 
we  have 

VA  =  VB, 


HYDRODYNAMICS  175 

and  consequently  the  above  equation  reduces  to 

—  =  —  +  (h  —  h)  =  —  -4-  h 
77  7 

The  pressure  head  at  entrance,  B,  is  therefore  simply  that  due 
to  the  static  head  hi  plus  that  due  to  the  atmospheric  pres- 
sure PA- 


.  158 


Similarly,  writing  Bernoulli's  equation  between  the  points  C 
and  D,  we  have 


whence 


PC  =  PD 
7  7 


The  effective  pressure  head  on  the  runner,  however,  is  the 
difference  in  pressure  between  the  entrance  B  and  the  exit  C; 
that  is 

Effective  head  =  —  —  — . 
7  7 

Substituting  the  values  just  obtained  for  pB  and  pc,  we  have, 
therefore, 


Effective  head  =    (—  -  ~  )   +  ^i  + 


176  ELEMENTS  OF  HYDRAULICS 

But  since  the  pressures  at  A  and  D  are  atmospheric,  we  have 
PA  —  PD-    Also  hi  +  h2  =  h.    Consequently  this  relation  becomes 

Effective  head  =  h  +  ^  ~  VD"\ 


This,  however,  neglects  all  frictional  losses  in  the  intake  and 


FIG.  159.— Single  horizontal  spiral  cased  turbine,  4000  H.  P.,  720  r.  p.  m., 
550  ft.  head.  Built  for  the  Cleveland  Cliffs  Iron  Co.,  Marquette,  Mich.,  by 
the  Allis  Chalmers  Manufacturing  Co. 

draft  tubes.  Including  such  losses,  the  actual  expression  for 
the  effective  head  becomes 

Effective  head  =  h  +  ^  ~  VD       -  friction  head. 

Provided  the  head,  h2,  in  the  draft  tube,  does  not  exceed  the 
ordinary  suction  head,  say  about  25  feet,  the  use  of  a  draft 
tube  therefore  produces  no  loss  in  the  static  head,  h.  Further- 
more, by  the  use  of  a  flaring  draft  tube  the  velocity  of  flow  at 


HYDRODYNAMICS 


177 


the  outlet  of  the  draft  tube  may  be  made  considerably  less  than 
that  at  the  inlet;  that  is,  for  a  flaring  tube,  VD  <  vc.     In  this  case 

the  term  -— ~ — -r  is  positive;  consequently  the  effective  head 


FIG.  160. — Vertical  shaft,  multi-runner  type  turbine,  showing  use  of  draft 
tubes.    S.  Morgan  Smith  Co. 

is  increased  by  this  amount,  and  is  therefore  greater  than  if 
no  draft  tube  was  used. 

Recent    Practice   in   Turbine    Setting. — A   typical    illustra- 
tion of  recent  development  in  the  design  and  installation  of 
12 


178 


ELEMENTS  OF  HYDRAULICS 


reaction  turbines  in  the  United  States  is  furnished  by  the  plant 
of  the  Mississippi  River  Power  Co.  at  Keokuk,  Iowa  (see  frontis- 
piece). These  turbines  are  of  the  Francis  type,  with  runners 
16  ft.  2  in.  in  nominal  diameter  and  develop  10,000  H.P.  per 
unit  at  a  speed  of  57.7  r.p.m.  under  a  head  of  32  ft.,  with  an  over- 
load capacity  of  13,000  H.P.  The  rated  efficiency  of  the  turbines 
is  88  per  cent. 

The  special  feature  which  makes  this  installation  typical  of 
recent  turbine  development  is  the  fact  that  each  of  the  30  units 


Head  Race 


FIG.  161. 

is  of  the  single  runner  vertical  shaft  type  as  shown  in  Fig.  162. 
The  trend  of  present  development  of  the  reaction  turbine  seems 
to  indicate  a  still  wider  application  of  this  type  to  all  conditions 
of  head  and  speed,  and  that  the  single  runner  vertical  shaft  tur- 
bine will  eventually  supersede  the  multi-runner  horizontal  shaft 
type  (Fig.  154)  and  the  multi-runner  vertical  shaft  type  (Fig. 
160). 

In  the  Keokuk  plant  the  intakes  and  draft  tubes  are  of  concrete 
moulded  in  the  substructure  of  the  power  house.  The  water 
from  the  forebay  reaches  each  turbine  through  four  intake  open- 
ings, the  outer  dimensions  of  which  are  22  ft.  by  7-1/2  ft.,  leading 
into  a  scroll  chamber  39  ft.  in  diameter  (Fig.  163).  The  draft 
tube  is  circular  and  18  ft.  in  diameter  immediately  below  the 
runner,  but  at  once  enlarges  and  changes  in  direction  from  vertical 


HYDRODYNAMICS 


179 


pOpoopoqc 
oooooboot 


FIG.  162. — Center  section  of  main  turbine  on  transverse  axis  of  power 
house.     Mississippi  River  Power  Co. 


180 


ELEMENTS  OF  HYDRAULICS 


I 

\ 

9 

OsF^'o^i 

/ 

OUTSIDE  UNMOfDRAfT  TUBt  BtNEMM 


HYDRODYNAMICS  181 

to  horizontal,  and  also  changes  in  cross  section  from  circular 
to  rectangular.  The  velocity  of  flow  is  thereby  diminished  from 
14  ft.  per  second  at  the  top  of  the  draft  tube  to  4  ft.  per  second 
at  the  outlet,  the  effect  being  to  increase  the  efficiency  about  7 
per  cent.  This  type  of  construction  is  also  representative  of 
recent  practice,  which  seems  to  favor  the  moulding  of  the  volute 
casing  directly  in  the  substructure  of  the  power  house  for  all 
low  head  work.  For  heads  exceeding  100  ft.,  the  amount  of 
reinforcement  in  the  concrete  becomes  so  great  as  to  warrant 
the  use  of  cast-iron  casings,  and  for  heads  exceeding  250  ft. 
the  use  of  cast  steel  for  turbine  casings  is  standard  practice. 

In  the  vertical  shaft  turbine  the  weight  is  carried  on  a  thrust 
bearing,  the  design  of  which  has  been  one  of  the  most  important 
considerations  affecting  the  adoption  of  this  type.  In  the 
Keokuk  plant  the  turbine  runner  is  coupled  to  the  generator  above 
by  a  shaft  25  in.  in  diameter  the  total  weight  of  the  revolving 
parts,  amounting  to  550,000  lb.,  being  carried  on  a  single 
thrust  bearing  6  ft.  in  diameter.  This  bearing  is  of  the  oil 
pressure  type,  a  thin  film  of  oil  being  maintained  at  a  pressure  of 
250  lb.  per  square  inch  between  the  faces  of  the  bearings.  As  a 
momentary  failure  of  the  oil  supply  would  result  in  the  immediate 
destruction  of  the  bearing,  provision  is  made  for  such  an  emergency 
by  introducing  an  auxiliary  roller  bearing  which  is  normally 
unloaded.  A  slight  decrease  in  the  oil  supply,  however,  allows 
the  weight  to  settle  on  this  roller  bearing,  which  although  not 
intended  for  permanent  use  is  sufficiently  large  to  carry  the 
weight  temporarily  until  the  turbine  can  be  shut  down. 

The  oil  pressure  bearing  when  taken  in  connection  with  the 
necessary  pumps  and  auxiliary  apparatus  is  expensive  to  install 
and  maintain,  and  requires  constant  inspection.  For  this  reason 
the  roller  bearing  and  the  Kingsbury  bearing  are  now  being 
applied  to  large  hydro-electric  units.  One  of  the  first  installations 
in  which  the  roller  bearing  was  applied  to  large  hydraulic  units 
was  at  the  McCall  Ferry  Plant  of  the  Pennsylvania  Water  and 
Power  Co.,  where  both  the  roller  and  the  Kingsbury  type  of 
bearing  are  now  in  satisfactory  use. 

36.    CHARACTERISTICS    OF   IMPULSE    WHEELS   AND    REACTION 

TURBINES 

Selection  of  Type. — The  design  of  hydraulic  turbines  is  a 
highly  specialized  branch  of  engineering,  employing  a  relatively 


182  ELEMENTS  OF  HYDRAULICS 

small  number  of  men,  and  is  therefore  outside  the  domain  of 
this  book.  On  account  of  the  rapid  increase  in  hydraulic  develop- 
ment, however,  every  engineer  should  have  a  general  knowledge 
of  turbine  construction  and  type  characteristics  so  as  to  be  able 
to  make  an  intelligent  selection  of  type  and  size  of  turbine  to  fit 
any  given  set  of  conditions.  For  this  reason  the  following  expla- 
nation is  given  of  the  use  and  significance  of  commercial  turbine 
constants,  such,  for  instance,  as  those  given  in  the  runner  table 
on  page  193. 

Action  and  Reaction  Wheels. — The  two  systems  of  hydraulic 
power  development  now  in  use  in  this  country  are  the  impulse 
wheel  and  the  radial  inward  flow  pressure  turbine.  When  an  im- 
pulse wheel  is  used,  the  total  effective  head  on  the  runner  is 
converted  into  speed  at  entrance  and  this  type  is  therefore 
sometimes  called  an  action  wheel.  In  the  case  of  a  pressure 
turbine,  however,  the  effective  head  on  the  runner  is  not  all 
converted  into  speed  at  entrance,  the  entrance  speed  being 
smaller  than  the  spouting  velocity,  so  that  the  water  flows 
through  the  runner  under  pressure,  the  effect  of  which  is  to 
accelerate  the  stream  as  it  passes  over  the  runner.  A  pressure 
turbine  is  therefore  called  a  reaction  wheel. 

Reaction  turbines  are  generally  used  for  heads  between  5 
and  500  ft.,  and  impulse  wheels  for  heads  between  about  300 
and  3000  ft.1  While  there  is  no  doubt  as  to  the  system  proper 
for  very  low  or  very  high  heads,  there  is  a  certain  intermediate 
range,  say  from  300  to  500  ft.,  for  which  it  is  not  directly  apparent 
which  system  is  most  suitable.  To  determine  the  proper  system 
within  this  range,  the  criterion  called  the  characteristic  speed 
has  been  introduced,  as  explained  in  what  follows. 

Speed  Criterion. — In  determining  the  various  criteria  for  speed, 
capacity,  etc.,  the  following  notation  will  be  used. 

Let  h  =  net  head  in  feet  at  turbine  casing, 

=  gross  head  minus  all  losses  in  head  race,  conduit  and 

tail  race; 

d  =  mean  entrance  diameter  of  runner  in  feet; 
b  =  height  of  guide  casing  in  feet; 

1  At  the  hydro-electric  plant  of  the  Georgia  Railway  and  Power  Co.  at 
Tallulah  Falls,  Georgia,  the  hydraulic  head  is  600  ft.,  which  is  probably  the 
highest  head  that  has  been  developed  east  of  the  Rocky  Mountains,  and 
the  highest  in  this  country  for  which  the  reaction  type  of  turbine  has  been 
employed.  (General  Electric  Review,  June  1914,  pp.  608-621.) 


HYDRODYNAMICS  183 

n  =  runner  speed  in  r.p.m.; 
v  =  spouting  velocity  in  feet  per  second; 
HI  =  peripheral  velocity  of  runner  in  feet  per  second; 

Ui  =   -'-  =  ratio  of  peripheral  speed  of  runner  to  spouting 
velocity  of  jet. 

From  Eq.  (23),  Art.  8,  the  spouting  velocity  in  terms  of  the 
head  is  given  by  the  relation 

v.= 

where  the  constant  C  =  0.96  to  0.97. 

For  maximum  efficiency  the  peripheral  velocity  of  the  runner  is 
some  definite  fraction  of  the  ideal  velocity  of  the  jet  \/2gh,  that  is, 

Ui   =   <£>V2gh  (lOO) 

where  <p  denotes  a  proper  fraction.  For  tangential  or  impulse 
wheels  the  average  value  of  v  is  from  0.45  to  0.51,  whereas  for 
reaction  turbines  its  value  ranges  from  0.49  to  0.96,  with  an 
average  range  from  about  0.57  to  0.87.1 

The  ratio  Ui  of  peripheral  speed  of  runner  to  spouting  velocity 
is  therefore  given  by  the  expression 

Ui          <f> 

Ul  ==  7  ==  C 

and  consequently  Ui  is  about  3  per  cent,  more  than  <p. 

Since  in  Eq.  (100)  the  factor  <p-\T2g  is  a  constant,  this  equa- 
tion may  be  written  in  the  form 


where  the  coefficient  kv  may  be  called  the  speed  constant.  For  a 
given  runner  for  which  d,  h  and  n  are  known,  this  speed  constant 
may  be  calculated  from  the  relation 

Ui  Trdn  ,       , 

k,  =  -j=  =     —--  (101) 

Vh 


1  For  the  numerical  values  of  these  and  other  constants  given  in  this 
article  see  the  periodical  literature  cited  in  succeeding  footnotes.  Also 
Gelpke  and  Van  Cleve,  Hydraulic  Turbines,  pp.  131-167;  Thurso,  Modern 
Turbine  Practice,  pp.  37-56;  Mead,  Water  Power  Engineering,  pp.  326-351; 
Daugherty,  Hydraulic  Turbines,  pp.  101-105. 


184  ELEMENTS  OF  HYDRAULICS 

By  the  use  of  the  speed  constant  kv,  different  types  of  runners 
may  be  compared  as  regards  speed.  In  the  case  of  reaction  tur- 
bines if  the  speed  constant  is  much  in  excess  of  7,  either  the  speed 
is  too  high  for  maximum  efficiency  or  the  nominal  diameter  of  the 
runner  is  larger  than  its  mean  diameter. 

Capacity  Criterion.  —  The  entrance  area  A  of  the  runner  is 
given  by  the  relation 

A 


where  Ci  denotes  a  proper  fraction,  since  the  open  circumference  is 
somewhat  less  than  the  total  circumference  by  reason  of  the  space 
occupied  by  the  ends  of  the  vanes  or  buckets.  The  velocity  of 
the  stream  normal  to  this  entrance  area  is  the  radial  component 
of  the  actual  .velocity  at  entrance,  say  ur,  and  like  this  velocity  is 
a  multiple  of  VA  ,  say 

Ur   = 


Since  the  discharge  Q  is  the  passage  area  multiplied  by  the  speed 
component  normal  to  this  area,  we  have 

Q  =  Aur  =  (ci7rc#>)c2V/i. 

It  is  customary,  however,  to  express  the  height  of  a  runner  in 
terms  of  its  diameter  as 

b  =  csd, 

where  the  coefficient  c3  is  a  constant  for  homologous  runners  of  a 
given  type.  For  American  reaction  turbines  c3  varies  from  about 
0.10  to  0.30.  Substituting  this  value  for  b  in  the  expression  for 
the  discharge,  it  becomes 

Q  =  (7rCiC2c3)d2V/i  . 

Therefore  if  the  constant  part  of  this  expression  is  denoted  by  kq, 
it  may  be  written 

Q  =  M2Vh:  (102) 

The  coefficient  kq  may  be  called  the  capacity  constant  of  the  run- 
ner, and  for  any  given  runner  may  be  computed  from  the  relation 


d'Vh 

For  American  reaction  turbines  the  capacity  constant  ranges  in 
value  from  about  2  to  4.     Since  kq  has  approximately  the  same 


HYDRODYNAMICS  185 

value  for  all  runners  of  a  given  type,  it  serves  as  a  criterion  for 
comparing  the  capacities  of  runners  of  different  types. 

Characteristic  Speed. — The  speed  constant  and  capacity 
constant  taken  separately  are  not  sufficient  to  fix  the  require- 
ments of  combined  speed  and  capacity.  That  is  to  say,  two 
runners  may  have  different  values  of  kv  and  kq  and  yet  be 
equivalent  in  operation.  To  fix  the  type,  therefore,  another 
criterion  must  be  introduced  which  shall  include  both  kv  and 
kq.  The  most  convenient  combination  of  these  constants  is 
that  introduced  by  Professor  Camerer  of  Munich  and  the  well- 
known  turbine  designer,  Mr.  N.  Baashuus  of  Toronto,  Ontario. 
This  criterion  may  be  obtained  as  follows.1 

The  horse  power  of  a  turbine  is  given  by  the  expression 

62.37QA6 
550 

where  e  denotes  the  hydraulic  efficiency  of  the  turbine.  If  the 
horse  power,  discharge  Q  and  head  h  are  given,  the  efficiency 
may  be  calculated  from  this  relation  by  writing  it  in  the  form 

550ff.P. 
~  62.37  Qh 

If  the  efficiency  is  known,  the  constants  in  the  above  expression 
for  the  horse  power  may  be  combined  into  a  single  constant  k, 
and  the  equation  written  in  the  form 

H.P.  =  kQh.  (104) 

When  the  efficiency  is  not  known  it  is  usually  assumed  as  80  per 
cent.,  in  which  case  k  =  yj- 

From  Eq.  (101)  the  speed  in  r.p.m.  is  given  by  the  relation 

6okvVh 


and  from  Eq.  (103)  the  nominal  diameter  of  the  runner  is  given  as 

Eliminating  d  between  these  two  relations,  we  have  therefore 

60  /bvV 


1  S.  G.  Zowski :  A  Comparison  of  American  Highspeed  Runners  for  Water 
Turbines.    Engineering  News,  Jan.  28,  1909,  pp.  99-102. 


186  ELEMENTS  OF  HYDRAULICS 

Moreover,  from  Eq.  (104)  we  have 

H.P. 


Q 


kh  ' 


and  substituting  this  value  of  Q  in  the  preceding  expression  for  n, 
we  have  finally 


n  =  ( -  - )  7~=-  (106) 

The  expression  in  parenthesis  is  a  constant  for  any  given  type  and 
may  be  denoted  by  Na,  in  which  case  we  have 


For  any  given  type  of  turbine  this  constant  Na  may  be  calculated 
from  the  relation 


n  VH.P. 

(I07) 


Various  names  have  been  proposed  for  this  constant  Ns  such 
as  "type  constant"  and  "type  characteristic."  In  Germany, 
where  its  importance  as  determining  the  type  and  performance 
of  a  turbine  seems  to  have  first  been  recognized,  it  is  called  the 
specific  speed  (spezifische  Geschwindigkeit,  or  spezifische  Um- 
laufzahl).  This  term,  however,  is  not  entirely  satisfactory  to 
American  practice,  as  it  seems  desirable  to  use  the  term  specific 
speed  in  another  connection,  as  explained  in  what  follows.  The 
term  for  the  constant  N8  favored  by  the  best  authorities  as  more 
fully  describing  its  meaning  is  characteristic  speed,  which  is  there- 
fore the  name  adopted  in  this  book.1 

For  impulse  wheels  the  characteristic  speed  ranges  in  value 
from  about  1  to  5,  while  for  radial,  inward  flow  turbines  its 
value  lies  between  10  and  100. 

Specific  Discharge.  —  It  is  convenient  to  express  the  discharge, 
power,  speed,  etc.,  in  terms  of  their  values  under  a  1  ft.  head. 

The  discharge  under  a  1  ft.  head  is  called  the  specific  dis- 
charge, and  its  value  is  found  by  substituting  h  =  I  in  Eq.  (102). 

1  The  use  of  the  term  characteristic  speed  has  been  recommended  to  the 
author  by  the  well  known  hydraulic  engineer  Mr.  W.  M.  White,  who  is 
using  this  term  in  preparing  the  American  edition  of  the  German  handbook 
"de  Hutte,"  and  strongly  advocates  its  general  adoption  in  American 
practice. 


HYDRODYNAMICS  187 

Consequently  if  the  specific  discharge  is  denoted  by  Qi,  its  value 
is 

Oi  =  M2, 

and  therefore 

Q  =  QiVh.  (108) 

For  reaction  turbines  the  specific  discharge  ranges  in  value  from 
0.302d2  for  the  slowest  speeds  to  2.866d2  for  the  highest  speeds, 
the  diameter  d  being  expressed  in  feet. 

Specific  Power. — Similarly,  the  power  developed  under  1  ft. 
head  is  called  the  specific  power  and  will  be  denoted  in  what  fol- 
lows by  H.P.i.  From  Eq.  (104)  we  have 

H.P.  =  kQh 

and  since  from  Eq.  (102) 

Q  =  M2  VJf , 

by  eliminating  Q  between  these  two  relations  we  have 
H.P.   =  kkgd2h^h. 

Substituting  h  =  1  in  this  equation,  the  specific  power  is  there- 
fore given  as 

H.P.!  =  kkqd\ 
and  consequently 

H.P.  =  ILP.ihVh.  (109) 

Specific  Speed. — By  analogy  with  what  precedes,  the  speed 
under  1  ft.  head  will  be  called  the  specific  speed  and  denoted 
by  HI.  Substituting  h  =  1  in  Eq.  (105),  we  have  therefore 

60/b, 

HI    =    -T-' 

ird 
and  consequently 

n  =  niVh.  (no) 

For  reaction  turbines  the  specific  speed  ranges  in  value  from 

78  147 

—r  for  the  lowest  speeds,  to  —j-  for  the  highest  speeds,  the  diameter 

d  being  expressed  in  feet. 
Relation  between  Characteristic  Speed  and  Specific  Speed. — 

From  the  relation 


188 


ELEMENTS  OF  HYDRAULICS 


the  characteristic  speed  Ns  may  be  defined  in  terms  of  the  quan- 
tities defined  above  as  specific.  Thus,  assuming  h  =  I  and 
H.P.  =  1,  we  have  Ns  =  n,  expressed  in  r.p.m.  Therefore, 
the  characteristic  speed  is  the  speed  in  r.p.m.  of  a  turbine  dimin- 
ished in  all  its  dimensions  to  such  an  extent  as  to  develop  1  H.P. 
when  working  under  a  head  of  1  ft. 

Since  it  is  apparent  from  Eq.  (106)  that  NB  stands  for  the 
combination 


where  A:  is  a  function  of  the  efficiency  e,  the  characteristic  speed 
Ns  is  an  absolute  criterion  of  turbine  performance  as  regards 
speed,  capacity  and  efficiency.  From  Eq.  (107),  however,  it 
is  evident  that  Ns  may  be  calculated  directly  from  the  speed, 
power  and  head  without  knowing  the  actual  dimensions  of 
the  runner,  its  discharge,  or  its  efficiency. 

Classification  of  Reaction  Turbines.  —  The  characteristic  speed 
N8  may  be  used  as  a  means  of  classifying  both  the  various 
types  of  impulse  wheels  and  of  reaction,  or  pressure,  turbines. 

In  the  following  table  practically  all  the  different  kinds  of 
pressure  turbines  of  the  radial  inward  flow  type  are  classified 
by  their  characteristic  speeds,  the  corresponding  efficiencies 
being  also  given  in  each  case.1 


Type  of  pressure 

Characteristic 

Efficiency 

turbine 

speed,  N» 

Maximum 

At  half  power 

Low  speed 

10  -  20 

82  per  cent,  at  3/4  power 

76  per  cent. 

Medium  speed.  .  . 
High  speed  

30  -  50 
60  -  80 

82  per  cent,  at  3/4  power 
80  per  cent,  at  0  .8  power 

75  per  cent. 
70  per  cent. 

Very  high  speed. 

90  -100 

73  per  cent,  at  0  .  9  power 

53  per  cent. 

The  values  of  the  constant  N8  in  this  table  refer  to  the  maximum 
power  of  one  runner  only.  In  case  the  characteristic  speed  is 
higher  than  100,  it  is  necessary  to  use  a  multiple  unit.  At 
maximum  power,  the  efficiencies  are  slightly  lower  than  the 
maximum  efficiencies  given  above. 

From  this  table  it  is  apparent  that  low  speed  turbines  show  a 
favorable  efficiency  over  a  wide  range  of  loads  but  are  prac- 
tically limited  to  high  heads,  whereas  high  speed  turbines  are 
efficient  at  about  0.8  load  but  show  a  notable  decrease  in  effi- 

1  Water  Power  Plants  and  the  Type  Characteristics  of  Turbines. 
N.  Baashuus,  Engineering  News,  Mar.  2,  1911,  pp.  248-250. 


HYDRODYNAMICS  189 

ciency  at  half  load.     The  use  of  the  latter  is  therefore  indicated 
for  low  heads  where  the  water  supply  is  ample  at  all  seasons. 

Classification  of  Impulse  Wheels. — In  a  way  similar  to  the 
preceding,  the  characteristic  speed  may  be  used  to  classify  the 
various  types  of  impulse  wheels,  as  indicated  in  the  following 
table.1 


Impulse  wheels 

Ng            |          1          |          2          |          3          |          4          |          5 

Efficiency   at 
3/4  load. 

80  per  cent. 

79  per  cent. 

78  per  cent. 

77  per  cent. 

76  per  cent. 

The  numerical  values  of  N8  in  this  table  refer  to  the  maximum 
power  of  one  nozzle  only.  In  case  the  characteristic  speed  lies 
between  5  and  10  it  is  therefore  necessary  to  use  more  than  one 
nozzle. 

Numerical  Applications. — To  illustrate  the  use  of  the  preceding 
numerical  data,  suppose  that  it  is  required  to  determine  the  proper 
system  of  hydraulic  development  for  a  power  site  with  an 
available  flow  of  310  cu.  ft.  per  second  under  an  effective  head  of 
324  ft. 

The  power  capacity  in  this  case  is 

Hp=  324X310  =9m  | 

Of  this  amount  about  100  H.P.  will  be  required  for  exciter  and 
lighting  purposes.  There  would  therefore  be  installed  two. 
exciter  units  running  at  550  r.p.m.,  one  of  which  would  be  a  re- 
serve unit.  The  characteristic  speed  for  these  units  would  then 
be 

550   /100 
N*  =  3lWl™  = 

Since  this  lies  between  1  and  5,  an  impulse  wheel  would  be  used 
for  driving  the  exciter  generators. 

The  main  development  of  9000  H.P.  would  be  divided  into  three 
units  of  3000  H.P.  each,  running  at  500  r.p.m.,  with  a  fourth 
unit  as  a  reserve.  The  characteristic  speed  for  these  main  units 
would  then  be 

500     /3000 
N*  ~  310  \17.6  = 

As  Ns  lies  between  10  and  100,  a  pressure  turbine  would  be  used 
for  driving  the  main  generators. 

1  Water  Power  Plants  and  the  Type  Characteristics  of  Turbines.  N. 
Baashuus,  Engineering  News,  Mar.  2,  1911,  pp.  248-250. 


190 


ELEMENTS  OF  HYDRAULICS 


As  a  second  illustration,  suppose  that  an  impulse  wheel  is 
required  to  develop  1300  H.P.  under  a  400  ft.  head  at  an  efficiency 
of  not  less  than  78  per  cent.  From  the  preceding  table  for  im- 
pulse wheels  it  is  apparent  that  it  is  necessary  to  use  a  wheel 
having  a  characteristic  speed  of  about  3.  If  a  single  wheel  and 
nozzle  is  used,  the  speed  in  r.p.m.  at  which  it  must  run  is  found 
from  Eq.  (107)  to  be 


n 


H.P. 


3  X  400 


150  r.p.m. 


If  two  nozzles  are  used,  each  furnishes  half  of  the  power  and  the 
corresponding  speed  is  150  V  2  =  212  r.p.m. 


perating 


Rai 


;        !  Overload 


10  20  30  40  50          60  70  80  90  100 

Per  Cent  Turbine  Load 

FlG.    164. 

With  four  nozzles  acting  on  two  runners  the  required  speed 
would  be  150V4  =  300  r.p.m.,  and  for  6  nozzles  acting  on  3 
runners,  n  =  150V6  =  367  r.p.m.  Since  the  value  of  N,  is  the 
same  in  each  case,  the  efficiency  is  practically  78  per  cent,  in 
each  case  although  there  is  a  wide  difference  in  the  speed  and 
setting. 

Normal  Operating  Range. — Having  determined  the  proper 
type  of  development,  it  is  necessary,  in  case  a  reaction  turbine 
is  used,  to  determine  the  required  size  and  type  of  runner  to 


HYDRODYNAMICS 


191 


develop  maximum  efficiency  under  the  given  conditions  of 
operation. 

For  a  turbine  direct  connected  to  a  generator,  the  capacity  of 
the  turbine,  in  general,  should  be  such  as  to  permit  the  full  over- 
load capacity  of  the  generator  to  be 
developed  and  at  the  same  time  place 
the  normal  operating  range  of  the  unit 
at  the  point  of  maximum  efficiency  of 
the  turbine,  as  indicated  in  Fig.  164. 
The  normal  horse  power,  or  full-load, 
here  means  the  power  at  which  the  max- 
imum efficiency  is  attained,  any  excess 
power  being  regarded  as  an  overload. 

When  the  supply  of  water  is  ample 
but  the  head  is  low,  efficiency  may  to  a 

certain  extent  be  sacrificed  to  speed  and  capacity  in  order  that 
the  greatest  power  may  be  developed  from  each  runner,  thereby 
reducing  the  investment  per  horse  power  of  the  installation.  On 
the  other  hand,  when  the  flow  of  water  is  insufficient  to  meet 
all  power  requirements,  an  increase  in  efficiency  shows  a  direct 
financial  return  in  the  increased  output  of  the  plant. 


FIG.  165. 


FIG.  166. 

Selection  of  Stock  Runner. — Ordinarily  it  is  required  to  select 
a  stock  runner  which  will  operate  most  favorably  under  the  given 
conditions.  To  explain  how  an  intelligent  selection  of  size  and 
type  of  runner  may  be  made  from  the  commercial  constants 
given  by  manufacturers,  the  following  runner  table  of  a  standard 
make  of  turbine  is  introduced.1  (Page  193.) 

1  The  Allis-Chalmers  Company,  Milwaukee. 


192 


ELEMENTS  OF  HYDRAULICS 


The  cut  accompanying  each  of  the  six  types  given  in  the  table 
shows  the  outline  of  runner  vane  for  this  type.  To  indicate  its 
relation  to  the  runner  and  to  the  turbine  unit  as  a  whole,  Fig. 


FIG.  168. 

165  shows  a  typical  cross  section  of  runner;  Fig.  166  shows  how 
this  is  related  to  the  casing;  and  Fig.  167  shows  a  cross  section 
of  the  entire  turbine  unit.  The  runner  is  also  shown  in  per- 
spective in  Fig.  168. 


HYDRODYNAMICS 


193 


TYPE  "A"  RUNNER 

nk 

TYPE"B"  RUNNER 

r' 

TYPE 

V  RUNNER 

-j-w- 

*  -" 

l> 

\ 

v  ^/ 

I 

IliV 

X 

Z 

JJL^S 

N 

5 

v» 

II 

4  

Na  '=  13.55 

Ns  =  20.3 

Na  = 

29.4 

Ui  - 

0.585 

Ui  =    0 

625 

Ui  = 

0.665 

Diam.   |  R.P.M.i 

H.P.i   |     Qi       R.P.M.i     H.P.i 

Ql 

RP.M.i     H.P 

.i 

Qi 

15          ^71.7 

0.0358 

0.394 

76.6 

0.0705 

0.776 

81.4 

0.130 

1.43 

18          "59.8 

0.0514 

0.565 

63.8 

0.105 

1.155 

67.8 

0.187. 

2.06 

21             51.2 

0.0705 

0.776 

54.7 

0.138 

1.523 

58.2 

0.225 

2.48 

24             44.8 

0.0915 

1.007 

47.8 

0.182 

2.00 

51.0 

0.333 

3.66 

27            39  .  8 

0.116 

1.276 

42.5 

0.229 

2.52 

45.2 

0.423 

4.65 

30            35.8 

0.142 

1.562 

38.3 

0.284 

3.12 

40.7 

0.520 

5.72 

34            31  _6 

0.184 

2.024 

33.8 

0.363 

3.99 

35.9 

0.668 

7.35 

38            28.3 

0.230 

2.53 

30.2 

0.453 

4.98 

32.2 

0.835 

9.19 

42            25.6 

0.280 

3.08 

27.4 

0.551 

6.06 

29.1 

1.016 

11.18 

46            23.4 

0.336 

3.69 

25.0 

0.665 

7.32 

26.6 

1.225 

13.48 

50            21.5 

0.398 

4.38 

23.0 

0.7b 

8.69 

24.4 

1.450 

15.95 

55             19.5 

0.480 

5.28 

20.9 

0.95 

10.45 

22.2 

1.745 

19.20 

60            17.9 

0.573 

6.30 

19.1 

1.13 

12.43 

20.4 

2.  OS 

\ 

22.88 

65             16.5 

0.672 

7.39 

17.7 

1.33 

14.63 

18.8 

2.44 

26.84 

70             15.4 

0.785 

8.64 

16.4 

1.53 

16.83 

17.5 

2.82 

31.00 

TYPE" 

D"  RUNNER 

j  .     j_ 

TYPE"E"  RUNNER 

TYPE"  F"  RUNNER 

^  —  V 

\ 

•h" 

> 

\» 

f) 

r 

' 

0 

a 

kr 

\ 

s^9- 

^ 

Q 

v  ^ 

' 

Na  =  40.7 

Ns  =  51.7—60 

5 

N8  =  71  . 

4—79 

Ui  - 

0.70 

Ui  =    0.75 

- 

Ui  =    0 

85 

Diam.|  R.P.M.i  |HJ>.i      Qi 

Diam.      R.P.M.i     H.P.i    I      Qi 

R.P.M.i 

H.P.i 

Qi 

14 

98.4 

0.277 

3.05 

111.5 

0.410 

4.51 

16 

86.1 

0.367 

4.04 

97.7 

0.541 

5.95 

18 

76.5 

0.471 

5.18 

86.8 

0.704 

7.74 

15 

85 

7 

0.226 

2.49 

20 

69.0 

0.597 

6.57 

78.1 

0.912 

10.03 

18 

71 

4 

0.324 

3.56 

22 

62.6 

0.731 

8.04 

71.0 

1.133 

12.46 

21 

61.3 

0.442 

4.86 

24 

57.4 

0.883 

9.70 

65.1 

1.375 

15.13 

24 

53.6 

0.577 

6.35 

26 

53.0 

.055 

11.60 

60.1 

1.62 

17.85 

27 

47 

6 

0.731 

8.04 

28 

49.2 

.243 

13.67 

55.8 

1.93 

21.25 

30 

42 

8 

0.902 

9.92 

30 

46.0 

.436 

15.80 

SS.l 

2.20 

24.20 

34 

37 

8 

1. 

158 

12.74 

32 

43.0 

.65 

18.15 

48.8 

2.55 

28.10 

38 

33 

9 

1.444 

15.88 

34 

40.5 

.8 

9 

20.80 

46.0 

2.82 

31.10 

42 

30 

6 

1.765 

19.4 

36 

38.3 

2.15 

23.65 

te.5 

3.14 

34.55 

46 

28 

0 

2.12 

23.3 

38 

36.3 

2.42 

26.60 

41.1 

3.52 

38.70 

50 

25 

7 

2.50 

27.5 

40 

34.4 

2.75 

30.25 

39.1 

3.93 

43.20 

55 

23 

4 

3.04 

33.4 

421/2 

32.4 

3.09 

34.0 

36.8 

4.33 

47.60 

60 

21 

4 

3.61 

39.7 

45 

30.6 

3.53 

38.8 

34.7 

4.92 

54.10 

65 

19.8 

4. 

22 

46.4 

47  1/2 

29.0 

4.01 

44.1 

32.9 

5.66" 

62.25 

70 

18 

4 

4.90 

53.9 

50 

27.6 

4.45 

49.0 

31.2 

6.13 

67.40 

52  1/2 

26.3 

4.95 

54.5 

29.8 

6.75 

74.25 

55 

25.1 

5.52 

60.7 

28.4 

7.50 

82.50 

57  1/2 

24.0 

6.10 

67.1 

27.2 

8.16 

89.75 

60 

23.0 

6.80 

74.8 

26.0 

8.94 

98.30 

13 


194  ELEMENTS  OF  HYDRAULICS 

From  Eq.  (107)  it  is  evident  that,  other  things  being  equal, 
the  characteristic  speed  for  high  heads  will  be  relatively  small 
whereas  for  low  heads  it  will  be  large.  Thus  in  the  runner 
table  above,  type  "A."  with  a  characteristic  speed  of  13.55  is 
adapted  to  high  heads,  running  up  to  600  ft.,  while  at  the  other 
end  of  the  series,  type  "F"  with  a  characteristic  speed  of  about 
75,  is  adapted  to  effective  heads  as  low  as  10  ft. 

To  give  a  numerical  illustration  of  the  use  of  the  runner  table, 
suppose  it  is  required  to  determine  the  type  of  runner  and  the 
speed  in  r.p.m.  to  develop  750  H.P.  under  a  head  of  49  ft. 

In  this  case  h^h  =  49V49  =  343,  and  consequently 

H.P.        750 
H-Rl-/*V/i     =343=   2'2> 

which  corresponds  to  a  30-in.  type  "F"  runner.  Referring 
to  the  table  for  this  type  and  size,  we  have  n\  =  52.1,  from  which 
the  required  speed  is  found  to  be 

n  =  52.  lV^  =  364.7,  say  360  r.p.m. 
If  twin  turbines  were  used,  we  would  have 

750 


which  corresponds  to  a  22-in.  type  "F"  runner,  having  a  speed 
of 

n  =  71.0Vft~  =  497,  say  500  r.p.m. 

As  a  second  illustration,  let  it  be  required  to  find  from  the 
table  the  type  of  runner  and  speed  to  develop  4000  H.P.  under  an 
effective  head  of  300  ft. 

In  this  case  h-Jh  =  300V300  =  5190,  and  consequently  the 
specific  power  is 

H.P.        4000 

H-Rl  ==  AV/T  :=  5190  =   °'77' 

which  corresponds   to   a   50-in.    type   "B"   runner.     Referring 
to  the  table  for  this  type  and  size  we  have 

H.P.i  =  0.79,  and  m  =  23, 
and  consequently  the  power  and  speed  for  this  type  and  size  is 

H.P.  =  0.79ft  V/T  =  4100, 
and 

n  =  23Y/T=  395,  say  400  r.p.m. 


HYDRODYNAMICS  195 

37.  POWER  TRANSMITTED  THROUGH  PIPE  LINE  AND  NOZZLE 

Effective  Head  at  Nozzle. — Hydraulic  power  is  frequently 
delivered  through  a  pipe  line  and  nozzle  (as,  for  instance,  when 
power  is  developed  by  an  impulse  wheel),  in  which  case  there  is 
considerable  loss  due  to  pipe  friction  and  other  causes.  To 
investigate  the  amount  of  this  loss  and  the  condition  for  maximum 
efficiency,  let 

D  =  inside  diameter  of  pipe  in  feet, 
I  =  length  of  pipe  in  feet, 
h  —  static  head  at  nozzle  in  feet, 
v  =  velocity  of  flow  in  feet  per  second. 

To  simplify  the  solution  it  is  customary  to  neglect  the  slight 
loss  of  head  at  entrance  to  the  pipe  and  in  the  nozzle,  as  both  of 
these  terms  are  small  in  comparison  with  the  head  lost  in  pipe  fric- 
tion. From  Eq.  (48),  Art.  17,  the  head  lost  in  friction  in  a 
pipe  of  length  I  is  given  by  the  expression 

I    v2 
Lost  friction  head  =  /  IV  o~~ 

where  /  denotes  an  empirical  constant  (Eq.  (49),  Art.  17,  and 
Table  12).  Therefore  the  effective  head  at  the  nozzle  is 

Effective  head  =  h  -  f^-  — •  (in) 

Velocity  of  Flow  for  Maximum  Power. — Since  the  discharge  Q 
is 

irD2 
Q  =  ~-A~vy  Ib.  per  second, 

the  horse  power  delivered  at  the  nozzle  is 

TrDVy      /,          ,1 
H.P.  at  nozzle  ==^T56   (h  ~  f 

The  value  of  v  for  which  the  horse  power  is  a  maximum  is  found 
from  the  calculus  condition 

d(H.P.) 

dv        =0' 
namety, 


dv        ~  4X550     V"         JD2g 
whence 

2ghD 
3fl    ' 


196  ,  ELEMENTS  OF  HYDRAULICS 

Maximum  Efficiency. — Substituting   this  value  of  v  in  the 
expression  for  the  horse  power,  its  maximum  value  is  found  to  be 


If  there  was  no  friction,  the  total  available  horse  power  would  be 

Total  H.P.  -  g^- 

The  efficiency  E  is  therefore 

_  actual  H.P.  J_  v^_ 

"total  H.P.  "  '  .X2)  V 

the  maximum  value  of  which,  when  v  =  -*  /  ^      ,  is 

I  2 

•"  wuzz.   power    ~    I  ~  ~~    =    ~T~'  (,^^4/ 

3         3 
Diameter  of  Nozzle  for  Maximum  Output  of  Power.--To 

investigate  conditions  at  the  nozzle,  let 

p  =  pressure  in  pounds  per  square  foot  before  entering  the 
nozzle, 

A  =  area  of  cross  section  of  pipe, 

a  =  area  of  cross  section  of  nozzle, 

v  =  velocity  of  water  in  pipe, 

V  =  velocity  of  water  through  nozzle. 

Then  since  all  the  water  in  the  pipe  discharges  through  the  nozzle, 
we  have  Av  =  aV,  or 

-       -, 

a  ~~  v 

and  therefore,  from  Bernoulli's  theorem, 

p        v2       V2        (A\ 2  v2 

~yJr2~g==~2g='  \a)    2g 
and  also 

2  J    4i2 

Therefore,  by  subtraction,  we  have 

/A\2v2  ,1  v2 

(a)    2g=h  ~  fD2g 
whence 

a  =  Av 


2ghD  -fW 


HYDRODYNAMICS 


197 


For  maximum  power  the  velocity  v,  as  given  by  Eq.  (112),  must 

be  v  =  +1  9      ,  and  inserting  this  value  of  v  in  the  expression  just 
\    oft 

obtained  for  the  nozzle  area  a,  the  condition  for  maximum  output 
of  power  is 

^  (115) 


For  a  circular  pipe  and  nozzle  we  have 
a 


Trd2        ,    ,         irD2 

-  and  A  =  -  -.— > 


where  d  denotes  the  nozzle  diameter;  hence  for  maximum  output 
we  must  have 


(116) 


30,000 


25,000 


20,000 


15,000 


10,000 


5,000 


100 


2 

60  .3 


40 


20 


10  15  20  25  30 

Velocity  in  Feet  per  Second 

FlG.    169. 


40 


45 


Graphical  Relation  between  Power  and  Efficiency. — The  rela- 
tion between  horse  power  and  efficiency  is  shown  graphically  in 
Fig.  169  for  the  installation  of  the  Vancouver  Power  Co.  The 
length  of  pipe  line  in  this  case  was  I  =  6678  ft.,  diameter  D  = 
4  ft.,  and  head  h  =  1214  ft.  Assuming  /  =0.024,  the  horse 
power  at  the  nozzle  is  found  from  the  equation 

H.P.  at  nozzle  =  ~^Q(h  -  f  5-^) 
=  1.42570(1214  -  0.622202). 


198 


ELEMENTS  OF  HYDRAULICS 


This  is  a  maximum  when 


" 


25.5, 


its  value  being  29,423  H.P. 

The  efficiency  in  per  cent,  is  found  from  the  relation 

E  =  (l  -  /^- 


=  (1  -  0.0005124^)100. 


The  horse  power  and  efficiency,  plotted  on  a  velocity  base,  are 
shown  in  Fig.  169. 

Method  of  Determining  Nozzle  Diameter.  —  In  finding  the 
nozzle  diameter,  d,  a  tentative  value  of  /  may  be  assumed  as  in 
the  example  just  cited  (see  Eq.  (48),  Art.  17),  and  the  cor- 
responding value  of  v  found  from  Eq.  (112).  The  value  of  / 
corresponding  to  this  velocity  may  then  be  determined  from 
Table  12,  and  with  this  value  of/,  the  nozzle  diameter  may  be 
determined  from  Eq.  (116). 

Note  that  the  nozzle  diameter  so  determined  gives  the  maximum 
output  of  power  for  the  size  of  pipe  specified,  but  that  more  power 
may  be  developed  from  the  same  nozzle  by  using  a  larger  supply 
pipe. 

38.  EFFECT  OF  TRANSLATION  AND  ROTATION 
Equilibrium  under  Horizontal  Linear  Acceleration.  —  Consider 
the  equilibrium  of  a  body  of  water  having  a  motion  of  translation 

as  a  whole  but  with  its  particles 
at  rest  relatively  to  one  another, 
such,  for  example,  as  the  water 
in  the  tank  of  a  locomotive 
tender  when  in  motion  on  a 
straight  level  track.  If  the 
speed  is  constant,  the  forces 
acting  on  any  particle  of  the 
liquid  are  in  equilibrium,  and 
conditions  are  the  same  as 
when  the  tank  is  at  rest.  If 
the  motion  is  accelerated,  how- 
ever, every  particle  of  the  liquid 
must  experience  an  inertia  force 
proportional  to  the  acceleration.  Thus,  if  the  acceleration  is 
denoted  by  a,  the  inertia  force  F  acting  on  any  particle  of  mass 
m,  according  to  Newton's  law  of  motion,  is  given  by  the  relation 

F  =  ma. 


a 
FIG.  170. 


HYDRODYNAMICS  199 

For  a  particle  on  the  free  surface  of  the  liquid  (Fig.  170),  the 
inertia  force  F  acting  on  this  particle  must  combine  with  its 
weight  W  into  a  resultant  R  having  a  direction  normal  to  the  free 
surface  of  the  liquid.  From  the  vector  triangle  shown  in  the 
figure  we  have 

F  =  W  tan  a, 
and  by  Newton's  law 


whence  by  division 

tan  a  =  —  (117) 

Equilibrium  under  Vertical  Linear  Acceleration.  —  If  the  tank 
is  moving  vertically  upward  or  downward,  the  surface  of  the 
liquid  will  remain  horizontal.  If  the  motion  is  uniform,  that 
is,  with  constant  velocity,  the  conditions  will  be  the  same  as 
though  the  tank  was  at  rest.  If  it  is  moving  upward  with 
acceleration  a,  the  surface  will  still  remain  horizontal  but  the 
pressure  on  the  bottom  of  the  tank  will  be  increased  by  the 

W 

amount  ma  =  —  a)  where  W  denotes  the  weight  of  a  column  of 

y 

water  of  unit  cross  section  and  height  equal  to  the  depth  of 
water  in  the  tank.     Thus  if  p  denotes  the  pressure  on  the  bottom 

W 

of  the  tank  in  pounds  per  square  inch,  then  since  ma  =  —  a  = 

yh 

—  a,  we  have 

("8) 


If  the  acceleration  is  vertically  downward,  the  pressure  on  the 
bottom  of  the  tank  is  diminished  by  the  amount  yh  (—)  ,    its 

*/ 

value  being 

(119) 


Free  Surface  of  Liquid  in  Rotation.  —  If  the  tank  is  in  the 
form  of  a  circular  cylinder  of  radius  r,  and  revolves  with  angular 
velocity  co  about  its  vertical  axis  Y  Y  (Fig.  171),  the  free  surface 
of  the  liquid  will  become  curved  or  dished.  To  find  the  form 
assumed  by  the  surface,  let  P  denote  any  particle  on  the  free 


200 


ELEMENTS  OF  HYDRAULICS 


surface  at  a  distance  x  from  the  axis  of  rotation.  Then  if  m 
denotes  the  mass  of  this  particle,  the  centrifugal  force  C  acting 
on  it  is 

n  9       W 

C  =  mx  co2  =  —  #co2. 

g 

From  the  vector  triangle  shown  in  the  figure  we  have 


W 


fj  n  I 

and  since  the  slope  of  the  surface  curve  is  ~  =  tan  0,  we  have 
as  its  differential  equation 


~  =  tan  0  = » 

do;  g 

whence,  by  integration,  its  algebraic  equation  is  found  to  be 

y  =  co^-  (120) 

The  surface  curve  cut  out  by  a  diametral  section  is  therefore 
a  parabola  with  vertex  in  the  axis  of  rotation,  and  the  free  surface 
is  a  paraboloid  of  revolution. 

Y 


E 


10 


FIG.  172. 


Depression  of  Cup  below  Original  Level  in  Open  Vessel. — 

Since  the  volume  of  a  paraboloid  is  half  the  volume  of  the 
circumscribing  cylinder,  the  volume  of  liquid  above  the  level 
OX  of  the  vertex  (Fig.  172)  is 

Vol.  OCDEF  = 


HYDRODYNAMICS 
But  if  AB  is  the  level  of  the  liquid  when  at  rest,  then 


201 


where  k  denotes  the  depth  of  the  cup  below  the  original  level, 
and  therefore 


2,.,2 


r'co 
4g 


(121) 


Consequently  the  depth  of  the  cup  below  the  original  level  is 
proportional  to  the  square  of 
the  angular  velocity. 

Depression  of  Cup  below 
Original  Level  in  Closed  Ves- 
sel.— If  the  top  of  the  tank  is 
closed  and  the  angular  velocity 
increased  until  the  liquid  presses 

against  the  top,   as  shown  in  J? 

Fig.  173,  the  surface  will  still     j  c 

remain   a   paraboloid.     If  the 

total  depth  of  the  cup  is  denoted 

by  H  and  its  greatest  radius  by 

R,  then  since  its  volume  must  be  the  same  as  that  of  the  cup 

of  depth  h,  we  have  the  relation 


FIG.  173. 


whence 


hr2  =  HR* 

^ 
tt     ''  H' 


,2-2 


But   from   the   equation  of   the  surface  curve,    y  =  ~^~,  by 
substituting  the  simultaneous  values  y  —  H,  x  =  R,  we  have 

H  =  °^-, 


and  substituting  in  this  relation  the  value  of  Rz  from  the  previous 
equation,  the  result  is 

H  =  °^> 
whence 

H  =  o>rA/^.  (122) 


202  ELEMENTS  OF  HYDRAULICS 

Therefore  after  the  liquid  touches  the  top  cover  of  the  tank,  the 
total  depth  of  the  cup  is  proportional  to  the  first  power  of  the 
angular  velocity. 

Practical  Applications. — An  important  physical  application 
of  these  results  consists  in  the  formation  of  a  true  parabolic 
mirror  by  placing  mercury  in  a  circular  vessel  which  is  then 
rotated  with  uniform  angular  velocity,  the  focus  of  the  mirror 
depending  on  the  speed  of  rotation. 

Another  practical  application  has  been  found  in  the  con- 
struction of  a  speed  indicator.  A  glass  cylinder  containing  a 
colored  liquid  is  mounted  on  a  vertical  spindle  which  is  geared 
to  the  shaft  whose  speed  is  required.  The  required  speed  is 
then  obtained  by  noting  the  position  of  the  vertex  of  the  para- 
boloid on  a  vertical  scale.  From  the  level  AB  to  the  level  CD 
(Fig.  173)  the  graduations  on  the  scale  are  at  unequal  distances 
apart,  as  apparent  from  Eq.  (121),  but  below  this  point  they  are 
equidistant,  as  shown  by  Eq.  (122). 

39.  WATER  HAMMER  IN  PIPES 

Increase   in  Pressure  due  to  Suddenly  Checking  Flow. — If 

water  is  flowing  in  a  pipe  with  uniform  velocity  and  the  flow  is 
suddenly  stopped,  as  by  the  closure  of  a  valve,  the  pressure  in 
the  pipe  is  greatly  increased.  If  the  velocity  of  flow  is  denoted 
by  v  and  the  mass  of  water  in  the  pipe  by  m,  the  momentum  of 
this  mass  will  be  mv.  Assuming  that  the  pipe  is  rigid  and  the 
water  incompressible,  if  the  flow  is  stopped  in  a  length  of  time  t, 
the  increased  pressure  F  near  the  valve  may  be  obtained  from 
the  principle  of  impulse  and  momentum,  namely 

Ft  =  mv.  (123) 

From  this  relation  it  appears  that  if  the  valve  is  closed  instan- 
taneously, that  is,  t  =  0,  the  pressure  F  is  infinite  since  the  right 
member,  mv,  is  finite  and  different  from  zero. 

Bulk  Modulus  of  Elasticity  of  Water. — In  reality,  however,  this 
result  is  not  valid,  since  the  pipe  walls  are  elastic  and  the  water 
compressible.  Thus  it  has  been  found  by  experiment  that  a  pres- 
sure of  one  atmosphere,  or  14.7  Ib.  per  square  inch,  on  each  face 
of  a  cube  of  water  at  32°  F.  causes  it  to  lose  about  0.00005  of  its 
original  volume.  Consequently  the  bulk  modulus  of  elasticity 
of  water,  B,  defined  as 

_  _     unit  Stress 

unit  volume  deformation 


HYDRODYNAMICS  203 

has  for  its  numerical  value 

B  =  -    — —  =  294,000  Ib.  per  square  inch.  (124) 

0.00005 

Pressure  Waves  in  Pipe  Produced  by  Suddenly  Checking 
Flow. — What  actually  happens  when  the  flow  in  a  pipe  is  suddenly 
shut  off,  is  an  increase  in  pressure,  beginning  at  the  valve,  which 
compresses  the  water  and  distends  the  pipe.  Beginning  at  the 
valve,  this  effect  travels  back  toward  the  reservoir  or  supply, 
producing  a  wave  of  compression  in  the  water  and  a  wave  of 
distortion  in  the  pipe.  When  all  the  water  in  the  pipe  has  been 
brought  to  rest,  the  total  kinetic  energy  originally  possessed  by 
the  flowing  water  is  stored  up  in  the  elastic  deformation  of  the 
water  and  pipe  walls.  Since  this  condition  cannot  be  maintained 
under  the  actual  head  in  the  reservoir,  the  pipe  then  begins  to 
contract  and  the  water  to  expand,  thereby  forcing  the  water 
back  into  the  reservoir  until  it  acquires  a  velocity  approximately 
equal  to  its  original  velocity  but  in  the  opposite  direction,  that  is, 
back  toward  the  reservoir.  After  this  wave  has  traversed  the 
pipe,  the  water  again  comes  to  rest,  but  the  kinetic  energy  ac- 
quired by  the  flow  toward  the  reservoir  will  have  reduced  the 
pressure  below  normal.  Consequently  water  again  enters  the 
pipe  from  the  reservoir  and  flows  toward  the  valve,  beginning  a 
new  cycle  of  operations. 

Velocity  of  Compression  Wave. — An  expression  for  the  velocity 
of  the  wave  of  compression  and  distortion  has  been  deduced  by 
Professor  I.  P.  Church,1  and  is  given  by  the  formula 


,        '--VicSrid)   '  <'*> 

where   vc  =  velocity  of  pressure  wave,  in  feet  per  second. 

B  =  bulk  modulus  of  elasticity  of  water  =  294,000  Ib. 

per  square  inch, 

E  =  modulus  of  elasticity  in  tension  of  pipe  material, 
c   =  thickness  of  pipe  wall  in  inches, 
d  =  internal  diameter  of  pipe  in  inches, 

8    =  -^r-r  =  0.4333. 
144 

Period  of  Compression  Wave. — The  period  of  the  pressure 
Church:  Hydraulic   Motors,  p.  203. 


204  ELEMENTS  OF  HYDRAULICS 

wave,  or  time  T  required  for  a  wave  to  make  a  complete  round 
trip  from  one  end  to  the  other  and  back  again,  is  evidently 

T  -  f-1,  (126) 

Vc 

where  I  denotes  the  length  of  the  pipe  in  feet. 

Increase  in  Pressure  Produced  by  Instantaneous  Stoppage  of 
Flow.  —  Professor  Church  has  also  derived  a  formula  for  the  in- 
crease in  pressure  due  to  an  instantaneous  stop,  namely, 

<5vvc  /      5BEa  (      , 

P~  =  -y    -  vVg(Ec+  BdT  (I27) 


g(Ec+ 

where  v  =  velocity  of  flow  before  closure, 

pmax  =  increase  in  pressure  in  pounds  per  square  inch  in  excess 
of  original  hydrostatic  pressure. 

Joukovsky  's  Experiments.  —  Elaborate  experiments  on  water 
hammer  in  pipes,  made  by  Professor  Joukovsky1  of  Moscow, 
Russia,  in  1897-98,  confirm  -these  theoretical  results,  and  also 
show  that  if  the  time  t  of  closure  is  less  than  the  period  T  (Eq. 
(126)),  the  full  force  of  the  maximum  pressure  (pmax,  Eq.  (127)) 
will  be  felt  in  the  pipe.  But  if  the  time  t  of  closure  is  greater  than 
the  period  T,  the  intensity  of  excess  pressure,  p,  is  reduced  ac- 
cording to  the  relation 


that  is,  the  excess  pressure  is  inversely  proportional  to  the  time  of 
closure. 

Gibson's  Experiments.  —  From  more  recent  experiments, 
made  by  A.  H.  Gibson2  of  Manchester,  England,  it  has  been 
found  that  when  the  time  of  stoppage  of  flow  is  greater  than  2  T 

4:1 

or  —  ,  the  excess  pressure  p  is  given  by  the  formula 

<-» 


where  a  =   cross  section  of  pipe, 

«i  =   area  through  valve  when  closing  begins, 
h  =   difference  in  static  head  in  feet  between  the  two  sides 

of  valve  when  there  is  no  flow, 
t  =   time  of  uniform  closing  of  valve. 

1  Joukovsky:  Abstract  by  Simin;  Jour.  Amer.  Water  Works  Assn.,  1904, 
p.  335. 

2  Gibson:  Water  Hammer  in  Hydraulic  Pipe  Lines,  Van  Nostrand,  1909. 


HYDRODYNAMICS 


205 


The  best  precaution  against  hydraulic  shock  of  this  nature  has 
been  found  to  be  the  use  of  slow  closing  valves.  Air  chambers 
placed  near  the  valves  have  also  been  found  effective  if  kept  filled 
with  air,  and  safety  valves  of  course  reduce  the  shock  to  a  pres- 
sure corresponding  to  the  strength  of  spring  used. 

40.  HYDRAULIC  RAM 

Principle  of  Operation.— A,  useful  application  of  water  hammer 
is  made  in  the  hydraulic  ram.  In  principle,  a  hydraulic  ram  is 
an  automatic  pump  by  which  the  water  hammer  produced  by  sud- 
denly checking  a  stream  of  running  water  is  used  to  force  a  portion 
of  that  water  to  a  higher  elevation. 


Drive  Pipe  Connection  A  Discharge  Pipe   F  Air  Chamber  E 

Escape  Valve    C  Delivery  Valve  D  Air  Feeder  H 

FIG.  174. — Rife  hydraulic  ram. 


To  illustrate  the  method  of  operation,  a  cross  section  of  a  ram 
is  shown  in  Fig.  174.  The  ram  is  located  below  the  level  of  the 
supply  water  in  order  to  obtain  a  flow  in  the  drive  pipe.  If  located 
some  distance  from  the  supply,  the  water  is  first  conducted  to  a 
short  standpipe,  as  shown  in  Fig.  175,  and  from  here  a  drive 
pipe  of  smaller  diameter  than  the  supply  pipe  conducts  the  water 
to  the  ram.  The  object  of  this  arrangement  is  to  utilize  the  full 


206 


ELEMENTS  OF  HYDRAULICS 


head  of  water  available  without  making  the  drive  pipe  too  long 
for  the  capacity  of  the  ram. 

Referring  to  Fig.  174,  the  water  flowing  in  the  drive  pipe  A  at 
first  escapes  around  the  valve  C,  which  is  open  or  down.  This 
permits  the  velocity  of  flow  to  increase  until  the  pressure  against 
C  becomes  sufficient  to  raise  it  against  its  seat  B.  Since  the 
water  can  then  no  longer  escape  through  the  valve  C,  it  enters  the 


Spring 


Strainer 


FIG.  175. 


air  chamber  E  through  the  valve  D,  thereby  increasing  the  pres- 
sure precisely  as  in  the  case  of  water  hammer  discussed  in  the 
preceding  article.  When  the  pressure  in  E  attains  a  certain  maxi- 
mum value,  the  flow  is  checked  and  the  valve  D  falls  back  into 
place,  closing  the  opening  and  trapping  the  water  which  has 
already  entered  the  chamber  E.  The  pressure  in  E  then  forces 
this  water  into  the  supply  pipe  F,  which  delivers  it  at  an  elevation 
proportional  to  the  pressure  in  E. 


Hydraulic  rams  are  also  so  built  that  they  can  be  operated  from 
one  source  of  supply  and  pump  water  from  a  different  source 
(Fig.  176).  Muddy  or  impure  water  from  a  creek  or  stream  may 
thus  be  used  to  drive  a  ram,  and  the  water  pumped  from  a  pure 
spring  to  the  delivery  tank. 

Efficiency  of  Ram. — The  mechanical  efficiency  of  a  ram  depends 
on  the  ratio  of  fall  to  pumping  head,  ranging  from  20  per  cent, 
for  a  ratio  of  1  to  30,  up  to  75  per  cent,  for  a  ratio  of  1  to  4. 


HYDRODYNAMICS  207 

Its  efficiency  as  a  pump  is  of  course  very  small,  as  only  a  small 
fraction  of  the  water  flowing  in  the  drive  pipe  reaches  the 
delivery  pipe.  The  advantages  of  the  hydraulic  ram  are  its 
small  first  cost,  simplicity  of  operation,  and  continuous  service 
day  and  night  without  any  attention. 

To  obtain  an  expression  for  the  mechanical  efficiency  of  a 
ram,  let 

H  =  supply  head, 

h  =  effective  delivery  head  including  friction, 
q  =  quantity  delivered, 
Q  =  quantity  wasted  at  valve. 

Then  the  total  input  of  energy  to  the  ram  is  (Q  +  q)H,  and 
the  total  output  is  qh.  Consequently  the  mechanical  efficiency 
is  given  by  the  ratio 

v  - 
~ 


This  is  known  as  d'Aubuisson's  efficiency  ratio. 

The  hydraulic  efficiency,  however,  is  the  ratio  of  the  energy 
required  for  delivery  to  the  energy  of  the  supply.  Consequently 
its  value  is 


QH 

The  latter  expression  is  known  as  Rankine's  formula. 

41.  DISPLACEMENT  PUMPS 

There  are  two  types  of  pumps  in  general  use;  the  displacement, 
or  reciprocating  type,  and  the  centrifugal  type.  In  the  displace- 
ment pump  the  liquid  is  raised  by  means  of  a  bucket,  piston,  or 
plunger,  which  reciprocates  backward  and  forward  inside  a 
cylindrical  tube  called  the  pump  barrel  or  cylinder.  In  the 
centrifugal  pump,  as  its  name  indicates,  the  operation  depends 
on  the  centrifugal  force  produced  by  rotation  of  the  liquid. 

Suction  Pump.  —  One  of  the  simplest  forms  of  displacement 
pump  is  the  ordinary  suction  pump  shown  in  Fig.  177.  Here 
the  essential  parts  are  a  cylinder  or  barrel  C  containing  a  bucket 
jB,  which  is  simply  a  piston  provided  with  a  movable  valve, 
permitting  the  water  to  pass  through  in  one  direction  only. 
This  bucket  is  made  to  reciprocate  up  and  down  inside  the  barrel 
by  means  of  a  rod  E.  A  suction  pipe  S  leads  from  the  lower 


208 


ELEMENTS  OF  HYDRAULICS 


end  of  the  barrel  to  the  liquid  to  be  raised,  and  a  delivery  pipe 
D  discharges  the  liquid  at  the  desired  elevation. 

In  operation,  the  bucket  starts  from  its  lowest  position,  and 
as  it  rises,  the  valve  m  closes  of  its  own  weight.  The  closing  of 
this  valve  prevents  the  air  from  entering  the  space  below  the 
bucket,  and  consequently  as  the  bucket  rises  the  increase  in 
volume  below  it  causes  the  air  confined  in  this  space  to  expand 

and  thereby  lose  in  pressure.  As  the 
pressure  inside  the  suction  pipe  S 
thus  becomes  less  than  atmospheric, 

••I j-  ^^5=^  the  pressure  outside  forces  some  of 

jo  E3-t=lr"D  ?  the  liquid  up  into  the  lower  end  of 

the  pipe. 

When  the  bucket  reaches  the  top 
of  its  stroke  and  starts  to  descend, 
the  valve  n  closes,  trapping  the  liquid 
already  in  the  suction  pipe  S  and  also 
that  in  the  barrel,  thereby  lifting  the 
valve  m  as  the  bucket  descends. 
When  the  bucket  reaches  its  lowest 
position,  it  again  rises,  repeating  the 
whole  cycle  of  operations.  At  each 
repetition  the  water  rises  higher  as 
it  replaces  the  air,  until  finally  it  fills 
the  pump  and  a  continuous  flow  is 
set  up  through  the  delivery  pipe. 

Maximum  Suction  Lift. — Since  atmospheric  pressure  at  sea 
level  is  14.7  Ib.  per  square  inch,  a  pump  operating  by  suction 
alone  cannot  raise  water  to  a  height  greater  than  the  head  cor- 
responding to  this  pressure.  Since  a  cubic  foot  of  water 
weighs  62.4  Ib.,  the  head  corresponding  to  a  pressure  of  one 
atmosphere  is 


FIG.  177. 


h  = 


62.4  "  0.434 
144 


ft'> 


which  is  therefore  the  maximum  theoretical  height  to  which 
water  can  be  lifted  by  suction  alone.  As  there  are  frictional  and 
other  losses  to  be  considered,  the  actual  suction  lift  of  pumps  is 
only  about  two- thirds  of  this  amount,  the  practical  lift  for  differ- 
ent attitudes  and  pressures  being  as  given  in  the  following  table. 


HYDRODYNAMICS 


209 


Altitude 

Barometric 
pressure 

Equivalent  head 
of  water 

Practical  suction 
lift  of  pumps 

Sea  level  .... 
1/4  mile.  .  . 
1/2  mile... 
3/4  mile... 
1    mile  

14.70  Ib.  per  sq.  in. 
14.02  Ib.  per  sq.  in. 
13.33  Ib.  per  sq.  in. 
12.661b.  per  sq.  in. 
12.02  Ib.  per  sq.  in. 

33.95ft. 
32.38ft. 
30.79ft. 

29.24ft. 
27.76ft. 

22ft. 
21ft. 
20ft. 
18ft. 
17ft. 

1-1/4  miles. 
1-1/2  miles. 
2    miles.  .  .  . 

11.42  Ib.  per  sq.  in. 
10.88  Ib.  per  sq.  in. 
9.881b.  per  sq.  in. 

26.38ft. 
25.13ft. 

22  .  82  ft. 

16ft. 
15  ft. 
14ft. 

Force  Pump. — When  it  is  necessary  to  pump  a  liquid  to  a 
height  greater  than  the  suction  lift,  or  when  it  is  desired  to  equalize 
the  work  between  the  up  and  down  strokes,  a  combination  suction 
and  force  pump  may  be  used,  as  shown  in  Fig.  178.  In  the 
simple  type  here  illustrated, 
the  bucket  is  replaced  by  a 
solid  piston,  the  movable 
valves  being  at  m  and  n  as 
shown.  On  the  up  stroke  of 
the  piston  the  valve  m  closes 
and  the  pump  operates  like  a 
simple  suction  pump,  filling 
the  barrel  with  liquid.  When 
the  piston  starts  to  descend, 
the  valve  n  closes,  and  the 
liquid  in  the  barrel  is  there- 
fore forced  out  through  the 
valve  m  into  the  delivery 
pipe  D. 

By  making  the  suction  and 
pressure  heads  equal,  the 
piston  can  therefore  be  made 
to  do  the  same  amount  of  work  on  the  down  as  on  the  up 
stroke;  or  the  entire  suction  head  may  be  utilized  and  the 
pressure  head  made  whatever  may  be  necessary. 

Stress  in  Pump  Rod. — To  find  the  pull  P  on  the  pump  rod  E 
for  the  type  shown  in  Fig.  177,  let  A  denote  the  area  of  the 
bucket  and  hi,  h2,  the  heads  above  and  below  the  bucket,  as 
indicated  in  the  figure.  Then  the  downward  pressure  PI  on  top 
of  the  bucket  is 


FIG.  178. 


Pi  =  14.74  +  62.4A2A, 


14 


210 


ELEMENTS  OF  HYDRAULICS 


and  the  upward  pressure  P2  on  the  bottom  of  the  bucket  is 

P2  =  14.7 A  -  62.4/hA. 
Therefore  the  total  pull  P  in  the  rod  is 

P  =  Pi  -  P*  =  62.4A(Ai  +  h*)  =  Q2AAh. 
If  I  denotes  the  length  of  the  stroke,  the  work  done  per  stroke  is 
then 

work  per  stroke  =  PI  =  62AAhl. 

For  the  combined  suction  and  pressure  type  shown  in  Fig. 
178,  the  pressure  in  the  rod  on  the  down  stroke  is 

P  =  62.4AA2, 

and  the  tension  in  the  rod  on  the  up  stroke  is 

P  =  Q2AAhi. 

Direct  Driven  Steam  Pump. — The  modern  form  of  recipro- 
cating power  pump  of  the  suction  and  pressure  type  is  the 


FIG.  179. 

direct  driven  steam  pump,  illustrated  in  Figs.  179  and  180.  In 
this  type  the  steam  and  water  pistons  are  on  opposite  ends  of 
the  same  piston  rod  and  therefore  both  have  the  same  stroke, 
although  their  diameters  are  usually  different.  Until  recently 
this  was  the  standard  type  of  general  service  pump,  being  used 


HYDRODYNAMICS 


211 


for  all  pressures  and  capacities,  from  boiler  feed  pumps  to  muni- 
cipal pumping  plants.  Although  the  centrifugal  type  is  rapidly 
taking  its  place  for  all  classes  of  service,  the  displacement  pump 
is  the  most  efficient  where  conditions  demand  small  capacity 
at  a  high  pressure,  as  in  the  operation  of  hydraulic  machinery. 
Fig.  181  illustrates  the  use  of  a  displacement  pump  in  connection 
with  a  hydraulic  press.  The  best  layout  in  this  case  would  be 


FIG.  180. 

to  use  a  high-pressure  pump  and  place  an  accumulator  (Art.  3) 
in  the  discharge  line  between  pump  and  press.  The  press  cylinder 
can  then  be  filled  immediately  at  the  maximum  pressure  and  the 
ram  raised  at  its  greatest  speed,  the  pump  running  meanwhile 
at  a  normal  speed  and  storing  excess  power  in  the  accumulator. 
Calculation  of  Pump  Sizes. — To  illustrate  the  calculation  of 
pump  sizes,  suppose  it  is  required  to  find  the  proper  size  for 
a  duplex  (i.e.,  two  cylinder,  Fig.  180)  boiler  feed  pump  to  supply 
a  100-H.P.  boiler. 


212 


ELEMENTS  OF  HYDRAULICS 


For  large  boilers  the  required  capacity  may  be  figured  as  34-1/4 
Ib.  of  water  evaporated  per  hour  per  horse  power.  For  small 
boilers  it  is  customary  to  take  a  larger  figure,  a  safe  practical 
rule  being  to  assume  1/10  of  a  gallon  per  minute  per  boiler  horse 
power. 


© 

^® 

©                      C 

@                       C 

©             ( 

)    <8 
)    <2 
5     <£ 

^ 

L- 

.  1 

|| 

HI 

rti 

,Kevcrse 
Valve 


To  Pressure  Governor 


Steam  Inlet 


Supply  to  I'russ  .^ 

Relief  Valve  o7scharge||  jil 


,Exhnnst 


FIG.  181. 

In  the  present  case,  therefore,  a  100-H.P.  boiler  would  require  a 
supply  of  10  gal.  per  minute.  Assuming  40  strokes  per  minute 
as  the  limit  for  boiler  feed  pumps,  the  required  capacity  is 

^Q  =  0.25  gal.  per  stroke. 

Therefore,  assuming  the  efficiency  of  the  pump  as  50  per  cent., 
the  total  capacity  of  the  pump  per  stroke  should  be 

0  25 

Q~  =  0.5  gal.  per  stroke. 

Since  we  are  figuring  on  a  duplex,  or  two  cylinder,  pump,  the 
required  capacity  per  cylinder  is 

-~  =  0.25  gal.  per  stroke  per  cylinder, 

z 

and  consequently  the  displacement  per  stroke  on  each  side  of  the 
piston  must  be 

0  25 

-~—  =  0.125  gal.  piston  displacement. 

A 


HYDRODYNAMICS  213 

Referring  to  Table  7  it  is  found  that  a  pump  having  a  water 
cylinder  2-3/4  in.  in  diameter,  with  a  5-in.  stroke,  will  have 
the  required  capacity. 

Power  Required  for  Operation. — To  find  expressions  for  the 
H.P.  and  steam  pressure  required  to  operate  a  displacement 
pump,  let 

Q  =  discharge  of  pump  in  gal.  per  minute, 

h   =  total  pumping  head  in  feet   (including  friction  and 

suction  head  if  any), 
D  =  diameter  of  steam  piston  in  inches, 
d  =  diameter  of  water  piston  in  inches, 
p  =  steam  pressure  in  pounds  per  square  inch, 
w  —  water  pressure  in  pounds  per  square  inch, 
n  =  number  of  full  strokes  (i.e.,  round  trips)  per  minute, 
c    =  number  of  pump  cylinders   (e.g.,  for  duplex  pump, 

Fig.  180,  c  =  2), 
I    =  length  of  stroke  in  inches, 
E  =  efficiency  of  pump. 

Since  a  gallon  of  water  weighs  8.328  lb.,  the  total  work  per 
minute  required  to  raise  the  given  amount  Q  to  the  height  h  is 

work  =  8.328Q/i  ft.-lb.  per  minute. 

Taking  into  account  the  efficiency  of  the  pump,  the  actual 
horse  power  required  is  therefore 

— •«*       <««> 

Diameter  of  Pump  Cylinder. — If  the  pump  makes  n  full 
strokes  per  minute,  the  piston  displacement  per  minute  for  each 
cylinder  is 


2n 


/7Td2\ 

far1) 


and  the  actual  effective  displacement  of  the  pump  per  minute  is 

/    fJ2  \ 

effective  displacement  =2ncE  l^TH  cu.  in.  per  minute. 

Equating  this  to  the  required  discharge  Q,  expressed  in  cubic 
inches  per  minute,  we  have 


~l     =  231Q, 


214  ELEMENTS  OF  HYDRAULICS 

whence  the  required  diameter  of  the  pump  cylinder  in  terms  of 
the  speed  is  found  to  be 


I2-13Vl^E  (13I) 

Steam  Pressure  Required  for  Operation. — Since  the  total 
pressure  on  the  steam  piston  cannot  be  less  than  that  on  the  water 
piston,  the  minimum  required  steam  pressure,  p,  is  given  by  the 
relation 

/7rZ>2\ 

p(-4-)  =  o, 

whence 

(132) 

Numerical  Application.— To  illustrate  the  application  of  these 
results,  suppose  it  is  required  to  determine  the  indicated  horse 
power  to  operate  a  fire  engine  which  delivers  two  streams  of  250 
gal.  per  minute  each,  to  an  effective  height  of  60  ft. 

Since  the  height  of  an  effective  fire  stream  is  approximately 
four-fifths  that  of  the  highest  drops  in  still  air,  the  required  head 
at  the  nozzle  is 

4/&  =  -^  X  60  =  75  ft. 

To  this  must  be  added  the  friction  head  hf  lost  in  the  hose 
between  pump  and  nozzle,  which  is  given  by  the  relation  (Art.  17) 


where  I  is  the  length  of  the  hose  and  d  its  diameter,  both  expressed 
in  inches,  v  is  the  velocity  of  flow  through  the  hose,  and/  is  an 
empirical  constant.  For  the  best  rubber  lined  hose,  /  =  0.02 
for  the  first  100  ft.  of  hose  and  0.0025  for  each  additional  100  ft. 
whereas  for  unlined  hose/  =  0.04  for  the  first  100  ft.  and  0.005 
for  each  additional  100  ft.  In  the  present  case,  assuming  100  ft. 
of  the  best  2-1/2-inch  rubber-lined  hose,  we  have  /  =  0.02,  and 
since  the  quantity  of  water  delivered  is 

231 
Q  =  250  X  Tyno  cu*  ^'  Per  minu^e> 

and  the  area  of  the  hose  is 

ird2       7r(2.5)2 
A  =  -j-  =  =  4.908  sq.  in., 


HYDRODYNAMICS  215 


^ 

the  velocity  of  flow  in  the  hose  is 


v  =  --  =  16.3  ft.  per  second. 


Consequently  the  friction  head  hf  is 


and  therefore  the  total  pumping  head  H  is 
H  =  75  +  40  =  115  ft. 

From  Eq.  (130)  the  total  horse  power  required,  assuming  a  pump 
efficiency  of  50  per  cent.,  is  then  found  to  be 

H.P.  =  0.00025  5°°0X50115  =  28.75. 

Assuming  the  efficiency  of  the  engine  to  be  60  per  cent.,  the  total 
indicated  horse  power  required  would  be 

28.75 
LILP-  ==  "060"  :     18' 

42.  CENTRIFUGAL  PUMPS 

Historical  Development.  —  The  centrifugal  pump  in  its  modern 
form  is  a  development  of  the  last  15  years  although  as  a  type  it  is 
by  no  means  new.  The  inventor  of  the  centrifugal  pump  was  the 
celebrated  French  engineer  Denis  Papin,  who  brought  out  the 
first  pump  of  this  type  in  Hesse,  Germany,  in  1703.  Another 
was  designed  by  Euler  in  1754.  These  were  regarded  as  curiosi- 
ties rather  than  practical  machines  until  the  type  known  as  the 
Massachusetts  pump  was  produced  in  the  United  States  in  1818. 
From  this  time  on,  gradual  improvements  were  made  in  the 
centrifugal  pump,  the  most  important  being  due  to  Andrews  in 
1839,  Bessemer  in  1845,  Appold  in  1848,  and  John  and  Henry 
Gynne  in  England  in  1851.  Experiments  seemed  to  show  that 
the  best  efficiency  obtainable  from  pumps  of  this  type  ranged  from 
46  to  64  per  cent,  under  heads  varying  from  4-1/2  to  15  ft.,  and  40 
ft.  was  considered  the  maximum  head  for  practical  operation. 

About  the  year  1901  it  was  shown  that  the  centrifugal  pump 
was  simply  a  water  turbine  reversed,  and  when  designed  on  similar 
lines  was  capable  of  handling  heads  as  large,  with  an  efficiency  as 


216  ELEMENTS  OF  HYDRAULICS 

high,  as  can  be  obtained  from  the  turbines  themselves.  Since 
this  date,  great  progress  has  been  made  in  both  design  and 
construction,  the  efficiency  of  centrifugal  pumps  now  ranging 
from  55  to  over  90  per  cent.,  and  it  being  possible  to  handle  heads 
as  high  as  300  ft.  with  a  single  stage-turbine  pump  and  prac- 
tically any  head  with  a  multi-stage  type.1 

The  advantages  of  the  centrifugal  over  the  displacement  type 
are  its  greater  smoothness  of  operation,  freedom  from  water 
hammer  or  shock,  absence  of  valves,  simplicity  and  compactness, 
and  its  adaptability  for  driving  by  belt  or  by  direct  connection  to 
modern  high-speed  prime  movers,  such  as  steam  turbines  and 
electric  motors.  Under  favorable  conditions  the  first  cost  of  a 
high-lift  centrifugal  pump  may  be  as  low  as  one-third  that  of  a 
displacement  pump,  and  the  floor  space  occupied  one-fourth  that 
required  by  the  latter.  However,  for  small  quantities  of  water 
discharged  under  a  high  head  the  displacement  pump  is  preferable 
to  the  centrifugal  type,  as  the  latter  requires  too  much  compound- 
ing under  such  conditions. 

Principle  of  Operation. — The  principle  on  which  the  original 
centrifugal  pumps  of  Papin  and  Euler  operated  was  simply 
that  when  water  is  set  in  rotation  by  a  paddle  wheel,  the  centrifu- 
gal force  created  forces  the  water  outward  from  the  center  of 
rotation.  Appold  discovered  that  the  efficiency  depended  chiefly 
on  the  form  of  the  blade  of  the  rotary  paddle  wheel,  or  impeller, 
and  the  shape  of  the  enveloping  case,  and  that  the  best  form  for 
the  blade  was  a  curved  surface  opening  in  the  opposite  direction 
to  that  in  which  the  impeller  revolved,  and  for  the  case  was  a 
spiral  form  or  volute.  The  first  engineer  to  discover  the  value  of 
compounding,  that  is,  leading  the  discharge  of  one  centrifugal 
pump  into  the  suction  of  another  similar  pump,  was  the  Swiss 
engineer  Sulzer  of  Winterthur,  who  was  closely  followed  by  A.  C. 
E.  Rateau  of  Paris,  France,  and  John  Richards  and  Byron  Jack- 
son of  San  Francisco,  California. 

In  its  modern  form,  the  power  applied  to  the  shaft  of  a  cen- 
trifugal pump  by  the  prime  mover  is  transmitted  to  the  water  by 
means  of  a  series  of  curved  vanes  radiating  outward  from  the 
center  and  mounted  together  so  as  to  form  a  single  member  called 

1  Rateau  found  by  experiment  that  with  a  single  impeller  3.15  in.  in 
diameter,  rotating  at  a  speed  of  18,000  r.p.m.,  it  was  possible  to  attain  a 
head  of  863  ft.  with  an  efficiency  of  approximately  60  per  cent.  Engineer, 
Mar.  7,  1902. 


HYDRODYNAMICS 


217 


Hollow  arm  impeller. 


Concave  arm  impeller. 


Sand  pump  impeller.  Open  impeller  used  in  sewage  pumps. 


Enclosed  side  suction  impeller.  Enclosed  double  suction  impeller. 

FIG.  182.— Impeller  types.    (Courtesy  Morris  Machine  Works.) 


218 


ELEMENTS  OF  HYDRAULICS 


the  impeller  (Fig.  182).  The  water  is  picked  up  at  the  inner 
edges  of  the  impeller  vanes  and  rapidly  accelerated  as  it  flows 
between  them,  until  when  it  reaches  the  outer  circumference  of 
the  impeller  it  has  absorbed  practically  all  the  energy  applied 
to  the  shaft. 

Impeller  Types. — There  are  two  general  forms  of  impeller, 
the  open  and  the  closed  types.  In  the  former  the  vanes  are 
attached  to  a  central  hub  but  are  open  at  the  sides,  revolving 
between  two  stationary  side  plates.  In  the  closed  type,  the 


FIG.  183. 

vanes  are  formed  between  two  circular  disks  forming  part  of 
the  impeller,  thus  forming  closed  passages  between  the  vanes, 
extending  from  the  inlet  opening  to  the  outer  periphery  of  the 
impeller.  The  friction  loss  with  an  open  impeller  is  considerably 
more  than  with  one  of  the  closed  type,  and  consequently  the 
design  of  pumps  of  high  efficiency  is  limited  to  the  latter. 

Conversion  of  Kinetic  Energy  into  Pressure. — As  the  water 
leaves  the  impeller  with  a  high  velocity,  its  kinetic  energy  forms 
a  considerable  part  of  the  total  energy  and  the  efficiency  of  the 
pump  therefore  depends  largely  on  the  extent  to  which  this 
kinetic  energy  is  converted  into  pressure  in  the  pump  casing. 

In  some  forms  of  pump  no  attempt  is  made  to  utilize  this 
kinetic  energy,  the  water  simply  discharging  into  a  concentric 


HYDRODYNAMICS 


219 


chamber  surrounding  the  impeller,  from  which  it  flows  into  a 
discharge  pipe.  The  result  of  such  an  arrangement  is  that  only 
the  pressure  generated  in  the  impeller  is  utilized  and  all  the  ki- 
netic energy  of  the  discharge  is  dissipated  in  shock  and  eddy 
formation. 

Volute  Casing. — This  loss  of  kinetic  energy  may  be  partially 
avoided  by  making  the  casing  spiral  in  section,  so  that  the 
sectional  area  of  the  discharge  passage  increases  uniformly, 


FIG.  184. — Double  suction  volute  pump,  built  by  the  Platt  Iron  Works  Co. 

making  the  velocity  of  flow  constant  (Fig.  183).  This  type  of 
casing  is  called  a  volute  chamber  (Fig.  184). 

When  the  volute  is  properly  designed,  a  high  efficiency  may 
be  obtained  with  this  type  of  casing.1 

Vortex  Chamber. — An  improvement  on  the  simple  volute 
chamber  is  that  known  as  the  whirlpool  chamber,  or  vortex 
chamber,  suggested  by  Professor  James  Thomson.  In  this  type 
the  impeller  discharges  into  a  concentric  chamber  considerably 
larger  than  the  impeller,  outside  of  and  encircling  which  is  a 
volute  chamber.  In  its  original  form  this  necessitated  exces- 

1With  the  De  Laval  volute  type  of  centrifugal  pump  shown  in  Fig.  185, 
efficiencies  as  high  as  85  per  cent,  have  been  obtained  under  favorable 
conditions. 


220 


ELEMENTS  OF  HYDRAULICS 


FIG.  185. — Longitudinal  section  of  De  Laval  single-stage  double-suction 

volute  pump. 


FIG.  186. — Longitudinal  section  of  Alberger  volute  pump. 


HYDRODYNAMICS 


221 


sively  large  dimensions,  but  in  a  modified  form  it  is  now  very 
generally  used  (Figs.  185  and  186). 

The  effectiveness  of  this  arrangement  depends  on  the  principle 
of  the  conservation  of  angular  momentum.  Thus,  after  the 
water  leaves  the  impeller  no  turning  moment  is  exerted  on  it 
(neglecting  frictional  resistance)  and  consequently  as  a  given 
mass  of  water  moves  outward,  its  speed  decreases  to  such  an 
extent  as  to  keep  its  angular  momentum  constant.  For  a  well- 
designed  vortex  chamber,  the  velocity  of  the  water  at  the  outside 


FIG.  187. — Diffusion  vanes. 

of  the  diffusion  space  is  less  than  the  velocity  of  the  water  as 
it  leaves  the  impeller  in  the  inverse  ratio  of  the  radii  of  these 
points,  and  if  this  ratio  is  large,  a  large  part  of  the  kinetic  energy 
of  the  discharge  may  therefore  be  converted  into  pressure  head  in 
this  manner.  This  method  of  diffusion  is  therefore  well  adapted 
to  the  small  impellers  of  high  speed  pumps,  since  the  ratio  of  the 
outer  radius  of  the  diffusion  chamber  to  the  outer  radius  of  the 
impeller  may  be  made  large  without  unduly  increasing  the  size  of 
the  casing. 

Diffusion  Vanes. — Another  method  for  converting  the  kinetic 
energy  of  discharge  into  pressure  head  consists  in  an  application 


222 


ELEMENTS  OF  HYDRAULICS 


of  Bernoulli's  law  as  illustrated  in  the  Venturi  tube;  namely,  that 
if  a  stream  flows  through  a  diverging  pipe  the  initial  velocity  head 
is  gradually  converted  into  pressure  head  without  appreciable 


END  SECTIONAL  VIEW 


SIDE  SECTJONAL  VIEW 

FIG.  188. — Alberger  turbine  pump. 

loss.  To  apply  this  principle  to  a  centrifugal  pump,  the  impeller 
is  surrounded  by  stationary  guide  vanes,  or  diffusion  vanes 
(Fig.  187),  so  designed  as  to  receive  the  water  without  shock 
on  leaving  the  impeller  and  conduct  it  by  gradually  diverging 
passages  into  a  vortex  chamber  or  volute  casing.  This  type  of 


HYDRODYNAMICS 


223 


construction  is  therefore  essentially  a  reversed  turbine,  and  is 
commonly  known  as  a  turbine  pump  (Fig.  188). 

The  angle  which  the  inner  tips  of  the  diffusion  vanes  make 
with  the  tangents  to  the  discharge  circle  is  calculated  exactly 
as  in  the  case  of  the  inlet  vanes  of  a  turbine,  that  is,  so  that  they 
shall  be  parallel  to  the  path  of  the  water  as  it  leaves  the  impeller. 
As  this  angle  changes  with  the  speed,  the  angle  which  is  correct 
for  one  speed  is  incorrect  for  any  other  and  may  actually  obstruct 
the  discharge.  A  turbine  pump  must  therefore  be  designed  for 
a  particular  speed  and  discharge,  and  when  required  to  work 
under  variable  conditions  loses  considerably  in  efficiency.  If 
the  conditions  are  very  variable,  the  vortex  chamber  type  is 
preferable,  both  by  reason  of  its  greater  average  efficiency  under 
such  conditions  and  also  on  account  of  its  greater  simplicity 
and  cheapness  of  construction. 


Discharge 


Suction 


FIG.  189. — Worthington  multi-stage  turbine  pump. 

Stage  Pumps. — Single  impellers  can  operate  efficiently  against 
heads  of  several  hundred  feet,  but  for  practical  reasons  it  is 
desirable  that  the  head  generated  by  a  single  impeller  should  not 
exceed  about  200  feet.  When  high  heads  are  to  be  handled, 
therefore,  it  is  customary  to  mount  two  or  more  impellers  on  the 
same  shaft  within  a  casing  so  constructed  that  the  water  flows 
successively  from  the  discharge  of  one  impeller  into  the  suction 


224  ELEMENTS  OF  HYDRAULICS 

of  the  next.  Such  an  arrangement  is  called  a  stage  pump, 
and  each  impeller,  or  stage,  raises  the  pressure  an  equal  amount. 
Fig.  189  shows  a  multi-stage  pump  of  the  turbine  type  and 
Fig.  190  one  of  the  volute  type. 

A  single  impeller  pump  may  be  either  of  the  side  suction  or 
double  suction  type.  In  the  latter,  half  of  the  flow  is  received 
on  each  side  of  the  impeller  which  is  therefore  perfectly  balanced 


im      m 


iHi 


FIG.  190. — De  Laval  multi-stage  volute  pump. 

against  end  thrust  (Fig.  185).  A  side  suction  pump,  however,  is 
simpler  in  construction,  and  it  is  also  possible  to  balance  them 
hydraulically  against  end  thrust  (Fig.  186).  In  stage  pumps  the 
device  sometimes  used  for  balancing  is  to  arrange  the  impellers 
in  pairs  so  that  the  end  thrust  of  one  impeller  is  balanced  by 
the  equal  and  opposite  end  thrust  of  its  mate. 

43.  PRESSURE  DEVELOPED  IN  CENTRIFUGAL  PUMP 

Pressure  Developed  in  Impeller. — The  pressure  produced  in  a 
centrifugal  pump  must  be  sufficient  to  balance  the  static  and 
frictional  heads.  When  there  is  no  volute,  vortex  chamber  or 
diffusor,  the  kinetic  energy  of  the  discharge  is  all  dissipated  and 
the  entire  change  in  pressure  is  produced  in  the  impeller.  If, 
however,  the  velocity  of  discharge  is  gradually  reduced  by  means 
of  one  of  these  devices,  a  further  increase  in  pressure  is  produced 
in  the  casing  or  diffusion  space,  and  if  a  diverging  discharge  pipe 
is  used  the  pressure  is  still  further  increased. 

The  change  in  pressure  which  is  produced  in  passing  through 
the  impeller  may  be  deduced  by  applying  Bernoulli's  theorem. 
For  this  purpose  it  is  convenient  to  separate  the  total  difference 


HYDRODYNAMICS 


225 


in  pressure  between  the  inlet  and  discharge  circles  into  two  com- 
ponents; one  due  to  the  rotation  of  the  water  in  a  forced  vortex 
with  angular  velocity  co,  and  the  other  due  to  the  outward  flow, 


FIG.  191. — Detail  of  labyrinth  rings  in  pump  shown  in  Fig.  190. 

£^ 


FIG.  192. 


i.e.,the  relative  motion  of  the  water  with  respect  to  the  vanes  of 
the  impeller.  Let  the  subscripts  1  and  2  refer  to  points  on  the 
inlet  and  discharge  circles  respectively.  Then  the  radii  of 


15 


226  ELEMENTS  OF  HYDRAULICS 

these  circles  will  be  denoted  by  r\,  r2;  the  pressure  at  any  point 
on  these  circles  by  pi,  p2,  etc.  Also  let  co  denote  the  angular 
velocity  of  rotation  of  the  impeller,  and  ui,  u2  the  tangential 
velocities  of  the  vanes  at  their  inner  and  outer  ends  (Fig.  192), 
in  which  case  Ui  =  r^u  and  Ui  =  r2o>. 

Applying  Bernoulli  theorem  to  the  change  in  pressure  produced 
by  rotation  alone,  we  have  therefore 

pi       w2n2  _  ^2       co2r22 
~  ~      ~~ 


Consequently  the  total  change  in  pressure  due  to  rotation,  say 
pr  where  pr  =  pz  —  PI,  is  given  by  the  relation 


y-        j  20V  2g 

This  expression  is  often  called  the  centrifugal  head. 
By  similar  reasoning  the  change  in  pressure  produced  by  the 
outward  flow  is  given  by  the  relation 

p\       wi2  _  p^      w<? 
whence 


T  g 

If  the  water  enters  radially,  </>i  =  90°  and  consequently  w^  = 
v\2  +  Ui2.  In  this  case,  denoting  the  difference  in  pressure  at 
inlet  and  exit  due  to  the  flow  by  pf,  where  pf  =  p'2  —  p'i,  we 
have 

Pf  =  P'Z  ~  P'1  =  ^i2  +  ^i2  ~  w-22 
T   "          T  2gr 

The  total  increase  in  pressure  in  the  impeller  between  the  inlet 
and  discharge  ends  of  the  vanes  is  therefore  given  by  the  relation 

pr  +  pf  _  vi2  +  ui2  -  wz2  +  u22  -  ui2      vi2  +  uz2  -  w22 
~ 


29  ^ 

Pressure  Developed  in  Diffusor.  —  Besides  the  increase  in 
pressure  produced  in  the  impeller,  the  use  of  a  suitable  diffusion 
chamber  permits  part  of  the  kinetic  energy  at  exit,  due  to  the 
absolute  velocity  v2  of  the  discharge  from  the  impeller,  to  be 
converted  into  pressure.  Thus  if  k  denotes  the  fraction  of  this 
kinetic  energy  which  is  converted  into  pressure  in  the  diffusor, 


HYDRODYNAMICS  227 

2 

the  expression  derived  above  is  increased  by  the  term  k^—     When 

diffusion  vanes  are  used,  as  in  a  turbine  pump,  the  value  of  k  may 
be  as  high  as  0.75,  and  for  a  vortex  chamber  it  may  reach  0.60. 

General  Formula  for  Pressure  Head  Developed. — Combining 
the  terms  derived  above,  the  total  pressure  head  H  developed  by 
the  pump  is  given  by  the  simple  formula 

kv22  +  Vi2  +  u22  —  w22. 
H  =  -  -^-  (133) 

In  applying  this  formula  it  is  convenient  to  note  that  the  total 
head  H  developed  in  the  pump  consists  of  three  terms,  as 
follows: 

^  =  head  at  eye  (entrance)  of  impeller; 
=  head  developed  in  impeller; 


— ^ —  =  head  developed  in  casing  or  diffusor. 

44.  CENTRIFUGAL  PUMP  CHARACTERISTICS 

Effect  of  Impeller  Design  on  Operation. — The  greatest  source  of 
loss  in  a  centrifugal  pump  is  that  due  to  the  loss  of  the  kinetic 
energy  of  the  discharge.  As  only  part  of  this  kinetic  energy  can 
be  recovered  at  most,  it  is  desirable  to  reduce  the  velocity  of 
discharge  to  as  low  a  value  as  is  compatible  with  efficiency  in 
other  directions.  This  may  be  accomplished  by  curving  the  outer 
tips  of  the  impeller  vanes  backward  so  as  to  make  the  discharge 
angle  less  than  90°.  The  relative  velocity  of  water  and  vane  at 
exit  has  then  a  tangential  component  acting  in  the  opposite  direc- 
tion to  the  peripheral  velocity  of  the  impeller,  which  therefore 
reduces  the  absolute  velocity  of  discharge.  This  is  apparent  from 
Fig.  193  in  which  the  parallelogram  of  velocities  in  each  of  the 
three  cases  is  drawn  for  the  same  peripheral  velocity  u2  and  radial 
velocity  at  exit  w2  sin  6%.  A  comparison  of  these  diagrams 
indicates" how  the  absolute  velocity  at  exit  v2  increases  as  the  angle 
62  increases.  The  backward  curvature  of  the  vanes  also  gives  the 
passages  a  more  uniform  cross  section,  which  is  favorable  to 
efficiency. 

The  effect  which  the  design  of  the  impeller  has  on  the  operation 
of  the  pump  is  most  easily  illustrated  and  understood  by  plotting 


228 


ELEMENTS  OF  HYDRAULICS 


curves  showing  the  relations  between  the  variables  under  con- 
sideration. Assuming  the  speed  to  be  constant,  which  is  the 
usual  condition  of  operation,  three  curves  are  necessary  to  com- 
pletely illustrate  the  operation  of  the  pump;  one  showing  the 
relation  between  capacity  and  head,  one  between  capacity  and 


Wz 


FIG.  193. 


power,  and  one  between  capacity  and  efficiency.     The  first  of 
these  curves  is  usually  termed  the  characteristic. 

Rising  and  Drooping  Characteristics. — The  principal  factor 
influencing  the  shape  of  the  characteristic  is  the  direction  of  the 


HYDRODYNAMICS 


229 


CAPACITY 

FIG.  195. — Characteristics  and  efficiency  curves  obtained  with   DeLaval 

pumps. 


230 


ELEMENTS  OF  HYDRAULICS 


tips  of  the  impeller  blades  at  exit,  although  there  are  other  fac- 
tors which  affect  this  somewhat.  If  the  tips  are  curved  forward 
in  the  direction  of  rotation  the  characteristic  tends  to  be  of  the 
rising  type,  whereas  if  they  curve  backward  the  characteristic 
tends  to  be  of  the  drooping  type  (Figs.  194  and  195) .  For  a  rising 
characteristic  the  head  increases  as  the  delivery  increases  and  con- 
sequently the  power  curve  also  rises,  since  a  greater  discharge 
against  a  higher  head  necessarily  requires  more  power  (Fig.  196). 
A  drooping  power  curve  may  be  obtained  by  throttling  at  the 
eye  of  the  impeller,  but  a  greater  efficiency  results  from  designing 
the  impeller  so  as  to  give  this  form  of  curve  normally. 


10   20   30  40   50   60   70   80   90  100  110  120  130  140  150  160  170 
Percentage  of  Normal  Capacity 

FIG.  196. 

For  a  high  lift  pump  under  an  approximately  constant  head,  as 
in  the  case  of  elevator  work,  a  pump  with  radial  vanes  is  most 
suitable  as  the  discharge  may  be  varied  with  a  small  altera- 
tion in  delivery  head.  This  is  also  true  for  a  pump  working 
under  a  falling  head,  as  in  the  case  of  emptying  a  lock  or  dry 
dock,  as  it  makes  it  possible  to  obtain  a  large  increase  in  the 
discharge  as  the  head  diminishes,  thereby  saving  time  although 
at  a  loss  of  efficiency. 

One  of  the  most  important  advantages  of  a  drooping  char- 
acteristic (Fig.  197)  is  that  it  is  favorable  to  a  drooping  power 
delivery  curve,  making  it  impossible  for  the  pump  to  overload 
the  driving  motor.  For  an  electrically  driven  pump,  in  which 
the  overload  is  limited  to  20  per  cent.,  or  at  most  25  per  cent.,  of 
the  normal  power,  backward  curved  vanes  are  therefore  essential. 
Moreover,  with  a  pump  designed  initially  to  work  against  a 


HYDRODYNAMICS 


231 


certain  head,  if  the  vanes  at  exit  are  radial,  or  curved  forward, 
the  possible  diminution  in  speed  is  very  small,  the  discharge 
ceasing  altogether  when  the  speed  falls  slightly  below  normal. 
As  the  backward  curvature  of  the  vanes  increases  the  range  of 
speed  also  increases,  and  consequently  when  the  actual  working 
head  is  not  constant,  as  in  irrigation  at  different  levels,  or  in 
delivering  cooling  water  to  jet  condensers  in  low  head  work, 
where  the  level  of  the  intake  varies  considerably,  a  pump  with 
drooping  characteristic  is  much  better  adapted  to  meet  varying 
conditions  without  serious  loss  of  efficiency. 


10   20   30   40   50   60   70   80   90   100  110  120  130  140  150  160 
Percentage  of  Normal  Capacity 

FIG.  197. 

Head  Developed  by  Pump. — These  facts  may  be  made  more 
apparent  by  the  use  of  the  expression  for  the  head  developed  by 
the  pump,  derived  in  the  preceding  article.  Considering  only 
the  head  developed  in  the  impeller  and  casing,  and  omitting  that 
due  to  the  velocity  of  flow  at  entrance,  t>i,  which  does  not  depend 
on  the  design  of  the  pump,  the  expression  for  the  head  developed 


s 


H  = 


Since  vz  is  the  geometric  resultant  of  u2  and  w2,  we  have  by  the 
law  of  cosines, 


COS 


For  an  ideal  pump,  that  is,  one  in  which  all  the  velocity  head  ~ 
is  converted  into  pressure  head  in  the  diffusor,  k  is  unity.     As- 


232  ELEMENTS  OF  HYDRAULICS 

suming  k  =  1  and  substituting  the  expression  for  #22  in  the  equa- 
tion for  H,  the  result  is 

W22  —  UvWz  COS  62 
Jn   =~  —  • 

g 

For  constant  speed  of  rotation,  u2  is  constant. 

For  forward  curved  vanes  02  is  greater  than  90°  and  therefore 
cos  B2  is  negative.  In  this  case  as  wz  increases  H  also  increases; 
i.e.,  the  greater  the  delivery  the  greater  the  head  developed. 

For  radial  tipped  vanes,  62  =  90°  and  cos  62  =  0.     In  this  case 

H  =  — ,  which  is  constant  for  all  deliveries. 

y 

For  backward  curved  vanes  62  is  less  than  90°  and  cos  62 
is  positive.  Consequently  in  this  case  as  the  delivery  increases 
the  head  diminishes. 

Although  these  relations  are  based  on  the  assumption  of  a 
perfect  pump,  they  serve  to  approximately  indicate  actual  condi- 
tions, as  is  evident  by  inspection  of  the  three  types  of  charac- 
teristic. 

Effect  of  Throttling  the  Discharge. — It  is  always  necessary 
to  make  sure  that  the  maximum  static  head  is  less  than  the  head 
developed  by  the  pump  at  no  discharge.  This  is  self-evident 
for  the  drooping  characteristic,  but  the  rising  characteristic 
is  misleading  in  this  respect  as  the  head  rises  above  that  at  shut- 
off.  Since  for  a  certain  range  of  head  two  different  outputs  are 
possible,  it  might  seem  that  the  operation  of  the  pump  under 
such  conditions  would  be  unstable.  This  instability,  however,  is 
counteracted  by  the  frictional  resistance  in  the  suction  and 
delivery  pipes,  which  usually  amounts  to  a  considerable  part  of 
the  total  head.  Any  centrifugal  pump  with  rising  characteristic 
will  therefore  work  satisfactorily  if  the  maximum  static  head 
is  less  than  the  head  produced  at  shut-off.  If  the  frictional 
resistance  is  small  it  may  be  increased  by  throttling  the  discharge, 
so  that  by  adjusting  the  throttle  it  is  possible  to  operate  the 
pump  at  any  point  of  the  curve  with  absolute  stability. 

Numerical  Illustration. — The  particular  curves  shown  in 
Fig.  198  were  plotted  for  an  8-in.,  3-stage  turbine  fire  pump 
built  by  the  Alberger  Co.,  New  York,  and  designed  to  deliver  750 
gal.  per  minute  against  an  effective  head  of  290  ft.,  the  pump 
being  direct  connected  to  a  75  H.P.  60-cycle  induction  motor 
operating  at  a  synchronous  speed  of  1200  r.p.m. 


HYDRODYNAMICS 


233 


The  head  curve  shows  that  this  pump  would  deliver  two  fire 
streams  of  250  gal.  per  minute  each  at  a  pressure  of  143  Ib. 
per  square  inch;  three  streams  of  250  gal.  per  minute  each  at  a 
pressure  of  125  Ib.  per  square  inch;  four  streams  of  250  gal.  per 
minute  each  at  a  pressure  of  nearly  100  Ib.  per  square  inch;  or 
even  five  fairly  good  streams  at  a  pressure  of  80  Ib.  per  square 
inch.  With  the  discharge  valve  closed  the  pump  delivers  no 
water  but  produces  a  pressure  equivalent  to  a  head  of  308  ft. 


420 

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0       100      200      300      400      500      600      700      800      900     1000    1100    1200     1300 
Gallonsiper  Minute 

FIG.  198. 

If  the  head  against  which  the  pump  operates  exceeds  this  amount, 
it  is  of  course  impossible  to  start  the  discharge.  The  head  for 
which  this  particular  pump  was  designed  was  290  ft.,  which  cor- 
responds to  the  point  of  maximum  efficiency.  It  is  therefore 
apparent  that  the  operating  head  must  be  carefully  ascertained  in 
advance,  for  if  it  is  higher  than  that  for  which  the  pump  was  de- 
signed, both  the  efficiency  and  the  capacity  are  diminished, 
whereas  if  it  is  lower,  the  capacity  is  increased  but  the  efficiency 
is  diminished. 


234  ELEMENTS  OF  HYDRAULICS 

The  power  curve  shows  that  under  low  heads  the  power  rises. 
Also  that  the  overload  in  the  present  case  is  confined  to  about 
12  per  cent,  of  the  normal  power.  Consequently  the  motor  could 
only  be  overloaded  12  per  cent,  if  all  the  hose  lines  should  burst, 
whereas  the  head  curve  shows  that  if  all  the  nozzles  were  shut 
off  no  injurious  pressure  would  result. 

The  efficiency  curve  always  starts  at  zero  with  zero  capacity, 
as  the  pump  does  no  useful  work  until  it  begins  to  discharge. 
The  desirable  features  of  an  efficiency  curve  are  steepness  at  the 
two  ends,  a  flat  top  and  a  large  area.  Steepness  at  the  beginning 
shows  that  the  efficiency  rises  rapidly  as  the  capacity  increases, 
whereas  a  flat  top  and  a  steep  ending  show  that  it  is  maintained 
at  a  high  value  over  a  wide  range.  Since  the  average  efficiency 
is  obtained  by  dividing  the  area  enclosed  by  the  length  of  the 
base,  it  is  apparent  that  the  greater  the  area  for  a  given  length, 
the  greater  will  be  the  average  efficiency. 


45.   EFFICIENCY  AND  DESIGN  OF  CENTRIFUGAL  PUMPS 

Essential  Features  of  Design. — The  design  of  centrifugal 
pumps  like  that  of  hydraulic  turbines  requires  practical  ex- 
perience as  well  as  detailed  mathematical  analysis.  The  general 
principles  of  design,  however,  are  simple  and  readily  understood, 
as  will  be  apparent  from  what  follows: 

Three  quantities  are  predetermined  at  the  outset.  The  inner 
radius  of  the  impeller,  riy  is  ordinarily  the  same  as  the  radius  of 
the  suction  pipe;  the  outer  radius,  r^  is  usually  made  twice  rijand 
the  angular  speed  co  at  which  the  impeller  is  designed  to  run  is 
fixed  by  the  particular  type  of  prime  mover  by  which  the  pump 
is  to  be  operated. 

The  chief  requirement  of  the  design  is  to  avoid  impact  losses. 
In  order  therefore  that  the  water  shall  glide  on  the  blades  of  the 
impeller  without  shock,  the  relative  velocity  of  water  at  entrance 
must  be  tangential  to  the  tips  of  the  vanes. 

Assuming  the  direction  of  flow  at  entrance  to  be  radial,  which 
is  the  assumption  usually  made  although  only  approximately 
realized  in  practice,  the  necessary  condition  for  entrance  with- 
out shock  is  (Fig.  192) 

Vi  =  HI  tan  0i, 


HYDRODYNAMICS  235 

which  determines  the  angle  0i.     The  relative  velocity  of  water 
and  vane  at  entrance  is  then 


The  direction  of  the  outer  tips  of  the  vanes,  or  angle  0%,  Fig. 
192,  is  determined  in  practice  by  the  purpose  for  which  the  pump 
is  designed,  as  indicated  in  Art.  44.  For  an  assigned  value 
of  62,  the  absolute  velocity  of  the  water  at  exit  is 


and  consequently  as  02  increases,  the  absolute  velocity  at  exit, 
t>2,  also  increases. 

Let  Sij  «2,  Fig.  192,  denote  the  radial  velocity  of  flow  at  en- 
trance and  exit,  respectively,  and  AI,  A2  the  circumferential 
areas  of  the  impeller  at  these  points.  Then  for  continuous 
flow  SiAi  =  SzA2.  Usually  si  =  $2,  in  which  case  A i  =  A2.  If 
61  and  62  denote  the  breadth  of  the  impeller  at  inlet  and  outlet 
respectively,  then  AI  =  2wr1bi  and  A2  =  27rr262,  and  conse- 
quently for  AI  =  Az  we  have  btfi  =  b2r2.  Assuming  the  radial 
velocity  of  flow  throughout  the  impeller  to  be  constant,  the 
breadth  b  at  any  radius  r  is  given  by  the  relation  br  =  bjTi. 

Hydraulic  and  Commercial  Efficiency. — Let  H'  denote  the 
total  effective  head  against  which  the  pump  operates,  including 
suction,  friction,  delivery  and  velocity  heads.  Then  if  w  denotes 
the  velocity  of  the  water  as  it  leaves  the  delivery  pipe,  h  the  total 
lift  including  suction  and  delivery  heads,  and  h/  the  friction  head, 
we  have 

H'  =  h  +  h,  +  |j. 

The  total  theoretical  head  H  developed  by  the  pump,  as  derived  in 
Art.  43,  is 

H=  — -. 


Consequently  the  hydraulic  efficiency  of  the  pump  is  the  ratio  of 
these  two  quantities,  that  is, 

jj/ 

Hydraulic  efficiency  =  •=-• 
n 

The  commercial  efficiency  of  the  pump  is  the  ratio  of  the  work 
actually  done  in  lifting  the  water  through  the  height  h  to  the 


236 


ELEMENTS  OF  HYDRAULICS 


total  work  expended  in  driving  the  impeller  shaft,  and  is  of  course 
less  than  the  hydraulic  efficiency. 

46.     CENTRIFUGAL  PUMP  APPLICATIONS 

Floating  Dry  Docks. — To  illustrate  the  wide  range  of  ap- 
plications to  which  centrifugal  pumps  are  adapted,  a  few  typical 
examples  of  their  use  will  be  given. 

The  rapid  extension  of  the  world's  commerce  in  recent  years  has 
created  a  demand  for  docking  facilities  in  comparatively  isolated 
ports,  which  has  given  rise  to  the  modern  floating  dry  dock 
(Fig.  199).  In  docks  of  this  type  the  various  compartments  into 


FIG.   199. 


which  they  are  divided  are  provided  with  separate  pumps  so  that 
they  may  be  emptied  in  accordance  with  the  distribution  of 
weight  on  the  dock.  Provision  is  usually  made  for  handling 
one  short  vessel,  two  short  vessels,  or  one  extremely  long  ship, 
the  balancing  of  the  dock  on  an  even  keel  being  accomplished  by 
emptying  the  various  compartments  in  proportion  to  the  weight 
sustained.  The  number  of  pumps  in  docks  of  this  type  varies 
from  6  to  20,  depending  on  the  number  of  compartments.  The 
centrifugal  pump  is  widely  used  and  particularly  suitable  for  this 
class  of  work,  where  a  large  quantity  of  water  has  to  be  discharged 
in  a  short  time  against  a  changing  head  which  varies  from  zero 


H  YD  ROD  YNA  MICS 


237 


when  the  pumping  begins  to  30  or  40  feet  when  the  dock  is  nearly 
dry  (Fig.  200). l 

Deep  Wells.— In  ob- 
taining a  water  supply 
from  deep  wells,  the 
problem  is  to  secure  a 
pump  which  will  handle 
a  large  quantity  of  water 
efficiently  in  a  drilled 
well  of  moderate  diam- 
eter, the  standard  di- 
ameters of  such  wells 
being  12  and  15  in.  To 
meet  this  demand,  cen- 
trifugal pumps  are  now 
built  which  will  deliver 
from  300  to  800  gal.  per 
minute  from  a  12-in. 
well,  and  from  800  to 
1500  gal.  per  minute 
from  a  15-in.  well,  with 
efficiencies  ranging  from 
55  to  75  per  cent.  The 
depth  from  which  the 
water  is  pumped  may 
be  300  ft.  or  more,  the 
pumps  being  built  in 
several  stages  according 
to  the  depth  (Fig.  201). 

Mine  Drainage. — The 
extensive  use  of  electric 
power  for  operating 
mining  machinery  has 
led  to  the  employment 
of  centrifugal  pumps  for 
mine  drainage.  The 
advantages  of  this  type 
of  pump  when  direct 
connected  to  a  high 


FIG.  200. 


1  Figs.    199-203  are  reproduced  by  permission  of  the  Platt  Iron  Works 
Co.,  Dayton,  Ohio. 


238 


ELEMENTS  OF  HYDRAULICS 


Deep    | 
Well  Centrif-    t 
ugal  Pump 


FiG.  201. 


FIG.  202. 


HYDRODYNAMICS 


239 


speed  motor  are  its  compactness,  simplicity  and  low  first  cost. 
Fig.  202  illustrates  a  mine  sinking  turbine  pump  which  operates 
against  a  1250  ft.  head  in  a  single  lift.  Pumps  of  similar  design 
are  in  operation  in  nearly  all  the  important  mining  regions  of 
the  United  States  and  Mexico.  The  turbine  pump  is  used  to  best 
advantage  where  it  is  required  to  unwater  a  flooded  mine  shaft. 
For  actual  sinking  work  a  displacement  pump  is  preferable 
unless  an  ample  sump  is  provided  in  order  to  keep  the  turbine 
pump  well  supplied  with  water  so  that  it  will  not  take  air. 

Fire  Pumps. — The  use  of  centrifugal  pumps  for  fire  protection 
has  been  formally  approved  by  the  Fire  Insurance  Underwriters, 


FIG.  203. 

who  have  issued  specifications  covering  the  essential  features  of 
a  pump  of  this  type  to  comply  with  their  requirements.  In 
the  case  of  fire  boats  the  centrifugal  pump  has  been  found  to 
fully  meet  all  demands.  The  New  York  fire  boats  "  James 
Duane"  and  "Thomas  Willett"  are  equipped  with  turbine 
pumps,  each  of  which  has  a  capacity  of  4500  gal.  per  minute 
against  150  Ib.  per  square  inch  pressure.  For  automobile  fire  en- 
gines, the  great  range  of  speed  for  gas  engines  gives  the  centrifugal 
pump  a  great  advantage,  making  it  possible  to  throw  streams 
to  a  great  height  by  merely  increasing  the  speed  of  the  motor. 
This  type  can  also  be  readily  mounted  on  a  light  chassis  and 
driven  from  the  driving  shaft  of  the  machine,  making  a  light, 
compact,  flexible  and  efficient  unit  (Fig.  203). 

Hydraulic  Dredging. — The  rapid  development  and  improve- 
ment of  internal  waterways  in  the  United  States  has  demonstrated 
the  efficiency  of  the  hydraulic  or  suction  dredge.  The  advantage 
of  the  hydraulic  dredge  over  the  dipper  and  ladder  types  is  that 
it  not  only  dredges  the  material  but  also  delivers  it  at  the  desired 


240  ELEMENTS  OF  HYDRAULICS 

point  with  one  operation.  Its  cost  for  a  given  capacity  is  also 
less  than  for  any  other  type  of  dredge,  while  its  capacity  is 
enormous,  some  of  the  Government  dredges  on  the  Mississippi 
handling  over  3000  cu.  yd.  of  material  per  hour. 

In  operation  the  dredging  pump  creates  a  partial  vacuum  in 
the  suction  pipe,  sufficient  to  draw  in  the  material  and  keep  it 
moving,  and  also  produces  the  pressure  necessary  to  force  the 
discharge  to  the  required  height  and  distance.  Hundreds  of 
such  pumps,  ranging  from  6  to  20  in.  in  diameter,  are  used 
on  western  rivers  for  dredging  sand  and  gravel  for  building  and 
other  purposes.  The  dredge  for  this  class  of  service  is  very  simple, 
consisting  principally  of  the  dredging  pump  with  its  driving 
equipment  mounted  on  a  scow,  the  suction  pipe  being  of  sufficient 
length  to  reach  to  the  bottom,  and  the  material  being  delivered 
into  a  flat  deck  scow  with  raised  sides,  so  that  the  sand  is  re- 
tained and  the  water  flows  overboard. 

For  general  dredging  service  where  hard  material  is  handled, 
it  is  necessary  to  use  an  agitator  or  cutter  to  loosen  the  material 
so  that  it  can  be  drawn  into  the  suction  pipe.  In  this  case  the 
suction  pipe  is  mounted  within  a  structural  steel  ladder  of 
heavy  proportions  to  stand  the  strain  of  dredging  in  hard 
material,  and  of  sufficient  length  to  reach  to  the  depth  required. 
The  cutter  is  provided  with  a  series  of  cutting  blades,  and  is 
mounted  on  a  heavy  shaft  supported  on  the  ladder,  and  driven 
through  gearing  by  a  separate  engine  (Fig.  204). 

Usually  two  spuds  are  arranged  in  the  stern  of  the  dredge  to 
act  as  anchors  and  hold  the  dredge  in  position.  The  dredge  is 
then  swung  from  side  to  side  on  the  spuds  as  pivots  by 'means  of 
lines  on  each  side  controlled  by  a  hoisting  engine,  thus  controlling 
the  operation  of  the  dredge. 

Suction  dredges  are  usually  equipped  with  either  12-,  15-,  18- 
or  20-in.  dredging  pumps,  the  last  named  being  the  standard 
size.  For  most  economical  operation  as  regards  power,  the 
velocity  through  the  pipe  line  should  not  be  greater  than  just 
sufficient  to  carry  the  material  satisfactorily. 

With  easily  handled  material  the  delivery  pipe  may  be  a  mile 
or  more  in  length,  but  with  heavy  mateiial  requiring  high  velocity 
the  length  should  not  exceed  4000  ft.  The  practical  maximum 
discharge  pressure  is  about  50  Ib.  per  square  inch.  For  long 
pipe  lines  it  therefore  becomes  necessary  to  use  relay  pumps, 
the  dredging  pump  delivering  through  a  certain  length  of  pipe 


HYDRODYNAMICS 


241 


10 


242  ELEMENTS  OF  HYDRAULICS 

into  the  suction  of  the  relay  pump,  and  the  latter  delivering  it 
through  the  remainder  of  the  line.  For  high  elevations  or  very 
long  lines,  several  relay  pumps  may  have  to  be  used. 

The  efficiency  of  a  dredging  pump  is  usually  only  40  or  50 
per  cent.,  a  high  efficiency  in  this  case  not  being  so  important 
as  the  ability  to  keep  going. 

Hydraulic  Mining. — The  centrifugal  pump  is  also  successfully 
used  in  hydraulic  mining,  where  a  high  pressure  jet  is  used  to 
wash  down  a  hill.  A  number  of  centrifugal  pumps  are  used  for 
this  purpose  in  the  phosphate  mines  of  Florida.  Other  uses  for 
centrifugal  pumps  besides  those  described  above  are  found  in 
municipal  water  works,  sewage  and  drainage  plants,  sugar  re- 
fineries, paper  mills  and  irrigation  works. 

APPLICATIONS 

101.  A  jet  2  in.  in  diameter  discharges  5  cu.  ft.  of  water  per 
second  which  impinges  on  a  flat  vane  moving  in  the  same  direc- 
tion as  the  jet  with  a  velocity  of  12  ft.  per  second.     Find  the  horse 
power  expended  on  the  vane. 

102.  A  fireman  holds  a  hose  from  which  a  jet  of  water  1  in. 
in  diameter  issues  at  a  velocity  of  80  ft.  per  second.     What  force 
will  the  fireman  have  to  exert  to  support  the  jet? 

103.  A  small  vessel  is  propelled  by  two  jets  each  9  in.  in  diam- 
eter.    The    water    is    taken    from  the  sea  through  a  vertical 
inlet  pipe  with  scoop  facing  forward,  and  driven  astern  by  a 
centrifugal  pump  2  ft.  6  in.  in  diameter  running  at  428  r.p.m. 
and  delivering  approximately  2250  cu.  ft.  of  water  per  second.     If 
the  speed  of  the  boat  is  12.6  knots  (1  knot  =  6080  ft.  per  hour), 
calculate  the  hydraulic  efficiency  of  the  jet. 

104.  In  the  preceding  problem,  the  efficiency  of  the  pump  was 
48  per  cent,  and  efficiency  of  engine  and  shafting  may  be  assumed 
as  80  per  cent.     Using  these  values,  calculate  the  total  hydraulic 
efficiency  of  this  system  of  propulsion. 

NOTE. — The  jet  propeller  is  more  efficient  than  the  screw  pro- 
peller, the  obstacle  preventing  the  adoption  of  this  system  in  the 
past  being  the  low  efficiency  obtainable  from  centrifugal  pumps. 

105.  A  locomotive  moving  at  60  miles  per  hour  scoops  up  water 
from  a  trough  between  the  rails  by  means  of  an  L-shaped  pipe 
with  the  horizontal  arm  projecting  forward.     If  the  trough  is 
2000  ft.  long,  the  pipe  10  in.  in  diameter,  the  opening  into  the 


HYDRODYNAMICS  243 

tank  8  ft.  above  the  mouth  of  the  scoop,  and  half  the  available 
head  is  lost  at  entrance,  find  how  many  gallons  of  water  are 
lifted  into  the  tank  in  going  a  distance  of  1600  ft.  Also  find  the 
slowest  speed  at  which  water  will  be  delivered  into  the  tank. 

106.  A  tangential  wheel  is  driven  by  two  jets  each  2  in.  in 
diameter  and  having  a  velocity  of  75  ft.  per  second.     Assuming 
the  wheel  efficiency  to  be  85  per  cent,  and  generator  efficiency  90 
per  cent.,  find  the  power  of  the  motor  in  kilowatts  (1  H.P.   =  746 
watts  =  0.746  kilowatt). 

107.  In  a  commercial  test  of  a  Pelton  wheel  the  diameter  of 
the  jet  was  found  to  be  1.89  in.,  static  head  on  runner  386.5 
ft.,  head  lost  in  pipe  friction  1.8  ft.,  and  discharge  2.819  cu.  ft. 
per  second.     The  power  developed  was  found  by  measurement  to 
be  107.4  H.P.      Calculate  the  efficiency  of  the  wheel. 

108.  A  nozzle  having  an  efflux  coefficient  of  0.8  delivers  a  jet 
1-1/2  in.  in  diameter.     Find  the  amount  and  velocity  of  the  dis- 
charge if  the  jet  exerts  a  pressure  of  200  Ib.  on  a  flat  surface 
normal  to  the  flow. 

109.  A  jet  2  in.  in  diameter  is  deflected  through   120°  by 
striking  a  stationary  vane.     Find  the  pressure  exerted  on  the 
vane  when  the  nozzle  is  discharging  10  cu.  ft.  per  second. 

110.  A  power  canal  is  50  ft.  wide  and  9  ft.  deep,  with  a  velocity 
of  flow  of  1-1/2  ft.  per  second.     It  supplies  water  to  the  turbines 
under  a  head  of  30  ft.     If  the  efficiency  of  the  turbines  is  80  per 
cent.,  find  the  horse  power  available. 

111.  It  is  proposed  to  supply  1200  electrical  horse  power  to 
a  city  25  miles  from  a  hydraulic  plant.     The  various  losses  are 
estimated  as  follows: 

Generating  machinery,  10  per  cent.;  line,  8  per  cent.;  trans- 
formers at  load  end,  9  per  cent. ;  turbine  efficiency,  80  per  cent. 

The  average  velocity  of  the  stream  is  3  ft.  per  second,  available 
width  90  ft.,  and  depth  6  ft.  Find  the  net  fall  required  at  the 
dam. 

112.  The  head  race  of  a  vertical  water  wheel  is  6  ft.  wide  and 

v 

the  water  9  in.  deep,  flowing  with  a  velocity  of  5  ft.  per  second.  If 
the  total  fall  is  20  ft.  and  the  efficiency  of  the  wheel  is  70  per 
cent.,  calculate  the  horse  power  available  from  it. 

113.  A  stream  is  150  ft.  wide  with  an  average  depth  of  4  ft. 
and  a  velocity  of  flow  of  1  ft.  per  second.     If  the  net  fall  at  the 
dam  is  20  ft.  and  the  efficiency  of  the  wheel  is  75  per  cent.,  find 
the  horse  power  available. 


244  ELEMENTS  OF  HYDRAULICS 

114.  Eighty  gallons  of  water  per  minute  are  to  be  pumped 
from  a  well  12  ft.  deep  by  a  pump  situated  50  ft.  from  the  well, 
and  delivered  to  a  tank  400  ft.  from  the  pump  and  at  80  ft. 
elevation.     The  suction  pipe  is  3  in.  in  diameter  and  has  two 
3-in.  elbows.     The  discharge  pipe  is  2-1/2  in.  in  diameter  and  has 
three  2-1/2-in.  elbows.     Find  the  size  of  engine  required. 

NOTE. — The  lift  is  92  ft.  and  the  friction  head  in  pipe  and 
elbows  amounts  to  about  25  ft.,  giving  a  total  pumping  head  of 
117  ft.  The  pump  friction  varies  greatly,  but  for  a  maximum 
may  be  assumed  as  50  per  cent,  of  the  total  head,  or,  in  the  present 
case,  58-1/2  ft. 

115.  A  single  acting  displacement  pump  raises  water  60  ft. 
through  a  pipe  line  1   mile  long.     The  inside  diameter  of  the 
pump  barrel  is  18  in.,  the  stroke  is  4  ft.,  and  the  piston  rod  is 
driven  by  a  connecting  rod  coupled  to  a  crank  which  makes  30 
r.p.m.     The  velocity  of  flow  in  the  pipe  line  is  3  ft.  per  second. 
Assuming  the  mechanical  efficiency  of  the  pump  to  be    75  per 
cent.,  and  the  slip  5  per  cent.,  find  the  horse  power  required  to 
drive  the  pump  and  the  quantity  of  water  delivered. 

116.  A  6-in.  centrifugal  pump  delivers  1050  gal.  per  minute, 
elevating  20  ft.     The  suction  and  discharge  pipes  are  each  6  in. 
in  diameter  and  have  a  combined  length  of  100  ft.     Find  the 
friction  head,  total  horse  power  required,  and  speed  of  pump 
for  50  per  cent,  efficiency. 

NOTE. — The  velocity  of  flow  in  this  case  is  12  ft.  per  second 
and  the  corresponding  friction  head  for  100  ft.  of  6-in.  pipe  is 
8.8  ft.  The  total  effective  head  is  therefore  28.8  ft.,  requiring 
15.26  H.P.  at  a  speed  of  410  r.p.m. 

117.  In  the  preceding  problem  show  that  if  an  8-in.  pipe  is 
used  instead  of  6-in.  there  will  be  a  saving  in  power  of  over  22 
per  cent. 

118.  A  hydraulic  ram  uses  1000  gal.  of  water  per  minute  under 
a  4-ft.  head  to  pump  40  gal.  per  minute  through  300  ft.  of  2-in. 
pipe  into  a  reservoir  at  an  elevation  of  50  ft.  above  the  ram. 
Calculate  the  mechanical  and  hydraulic  efficiencies  of  the  ram, 
assuming  the  coefficient  of  pipe  friction  as  0.024. 

119.  An  automobile  booster  fire  pump,   used  for  making  a 
quick  initial  attack  on  a  fire,  is  required  to  deliver  two  streams 
through  3/8-in.  nozzles  and  250  ft.  of  1-in.  hose.     The  pump  is  of 
the  centrifugal  type  and  is  geared  up  to  a  speed  of  3500  r.p.m. 
from  the  gas  engine  which  drives  the  machine.     Calculate  the 


HYDRODYNAMICS  245 

discharge  in  gallons  per  minute  and  the  horse  power  required  to 
drive  the  pump,  assuming  50  per  cent,  efficiency. 

NOTE. — For  this  size  nozzle,  the  maximum  discharge  is  reached 
with  a  nozzle  pressure  of  about  68  Ib.  per  square  inch  corre- 
sponding to  a  velocity  of  about  100  ft.  per  second. 

120.  Feed  water  is  pumped  into  a  boiler  from  a  round  vertical 
tank  2-1/2  ft.  in  diameter.     Before  starting  the  pump  the  water 
level  in  the  boiler  is  38  in.,  and  in  the  tank  22  in.,  above  the  floor 
level,  and  when  the  pump  is  stopped  these  levels  are  40  in.  and 
15  in.  respectively.     If  the  steam  pressure  in  the  boiler  while  the 
pump  is  at  work  is  100  Ib.  per  square  inch  find  the  number  of 
foot-pounds   of  work  done  by  the  pump. 

121.  A  fire  pump  delivers  three  fire  streams,  each  discharging 
250  gal.  per  minute  under  80  Ib.  per  square  inch  pressure.   Find 
the  horse  power  of  the  engine  driving  the  pump  if  the  efficiency 
of  the  engine  is  70  per  cent,  and  of  the  pump  is  60  per  cent. 

122.  A  mine  shaft  580  ft.  deep  and  8  ft.  in  diameter  is  full  of 
water.     How  long  will  it  take  a  6-H.P.  engine  to  unwater  the 
shaft  if  the  efficiencies   of  pump  and   engine  are  each  75   per 
cent.? 

123.  A  fire  engine  pumps  at  the  rate  of  500  gal.  per  minute 
against  a  pressure  of  100  Ib.   per  square  inch.     Assuming  the 
overall  efficiency  to  be  50  per  cent.,  calculate  the  indicated  horse 
power  of  the  engine. 

124.  A  water  power  plant  is  equipped  with  tangential  wheels 
having  an  efficiency  of  80  per  cent.     The  water  is  delivered  to 
the  wheels  through  a  cylindrical  riveted  steel  penstock  5  miles 
long  with  a  total  fall  of  900  ft.,  practically  the  entire  penstock 
being  under  this  head. 

The  cost  of  power  house  and  equipment  is  estimated  at  $50,000, 
penstock  6  cents  per  pound,  operating  expenses  $5000  per  annum, 
and  interest  on  total  investment  4  per  cent,  per  annum.  The 
income  is  to  be  derived  from  the  sale  of  power  at  $12  per  horse 
power  per  annum.  A  constant  supply  of  water  of  100  cu.  ft. 
per  minute  is  available. 

(A)  Plot  a  curve  with  diameter  of  penstock  as  abscissa  and 
yearly  gross  income  as  ordinate. 

(B)  Plot  a  curve  with  diameter  of  penstock  as  abscissa  and 
yearly  gross  expenses  as  ordinate. 

(C)  From  these  two  curves  determine  the  two  values  of  the 
diameter  of  penstock  for  either  of  which  the  net  income  is  zero, 


246 


ELEMENTS  OF  HYDRAULICS 


and  the  one  value  for  which  the  net  income  is  a  maximum,  and 
find  the  amount  of  the  latter. 

125.  A  hydraulic  pipe  line  is  required  to  transmit  150  H.P. 
with  a  velocity  of  flow  not  greater  than  3  ft.  per  second  and  a 
delivery  pressure  of  900  Ib.  per  square  inch.     Assuming  that 
the  most  economical  size  of  pipe  is  one  which  allows  a  pres- 
sure drop  of  about  10  Ib.  per  square  inch  per  mile,  determine 
the  required  size  of  pipe. 

126.  Find  the  maximum  horse  power  which  can  be  transmitted 
through  a  6-in.  pipe  4  miles  long  assuming  the  inlet  pressure  to 
be  800  Ib.  per  square  inch  and  the  coefficient  of  pipe  friction 
to  be  0.024.     Also  determine  the  velocity  of  flow  and   outlet 
pressure. 

127.  A  6-in.  pipe  half  a  mile   long  leads   from   a   reservoir 
to  a  nozzle  located  350  ft.  below  the  level  of  the  reservoir  and 
discharging  into  the  air.     Assuming  the  coefficient  of  friction  to 
be  0.03,  determine  the  diameter  of  nozzle  for  maximum  power. 

128.  A  10-in.  water  main  900  ft.  long  is  discharging  1000  gal. 
of  water  per  minute.     If  water  is  shut  off  in  2  sec.  by  closing 
a  valve,  how  much  is  the  pressure  in  the  pipe  increased? 

129.  In   a   series    of   experiments    made    by   Joukowsky    on 
cast-iron  pipes,  the  time  of  valve  closure  in   each    case   being 
0.03  sec.,  the  following  rises  in  pressures  were  observed.1     Show 
that  these  results  give  the  straight  line  formula,  p  =  57v. 


CAST-IRON  PIPE,  DIAM.  4  IN.,  LENGTH  1050  FT. 


Vel.  in  ft.  /sec  

0.5 

2.0 

3.0 

4.0 

9.0 

Observed  pressure  in  lb./in.2.  .  . 

31 

119 

172 

228 

511 

Cast-iron  pipe,  diameter  6  in.,  length  1066  ft. 

Vel  in  ft  /sec                                          .  .  . 

0.6 

2.0 

3.0 

7.5 

Observed  rise  in  pressure  .in  lb./in.2.  .  .  . 

43 

113 

173 

426 

130.  It  is  customary  in  practice  to  make  allowance  for  possible 
water  hammer  by  designing  the  pipes  to  withstand  a  pressure  of 
100  Ib.  per  square  inch  in  excess  of  that  due  to  the    static 
head.     Show  that  this  virtually   allows   for   an   instantaneous 
stoppage  at  a  velocity  of  1.6  ft.  per  second. 

131.  A  bowl  in  the  form  of  a  hemisphere,  with  horizontal  rim, 
Gibson:    Hydraulics  and  Its  Applications,  p.  239. 


HYDRODYNAMICS 


247 


is  filled  with  liquid  and  then  given  an  angular  velocity  co  about 
its  vertical  axis.  How  much  liquid  flows  over  the  rim  (Fig. 
205)? 

132.  A  closed  cylindrical  vessel  of  height  H  is  three-fourths 
full  of  water.     With  what  angular  velocity  co  must  it  revolve 
around  its  vertical  axis  in  order  that  the  surface  paraboloid 
shall  just  touch  the  bottom  of  the  vessel 

(Fig.  206). 

133.  A    closed    cylindrical    vessel    of 
diameter  3  ft.  and  height  6  in.  contains 
water  to  a  depth  of  2  in.      Find  the 
speed  in  r.p.m.  at  which  it  must  revolve 
about  its  vertical  axis  in  order  that  the 
water  shall  assume  the  form  of  a  hollow 
truncated    paraboloid    for    which     the 
radius  of  the  upper  base  is  1  per  cent, 
greater   than   the  radius  of  the   lower 

base;  or,  referring  to  Fig.  207,  such  that  r\  =  1.01r2. 

134.  The  test  data  for  a  19-in.  New  American  turbine  runner 
are  as  follows: 

Head  25  ft.;  speed  339  r.p.m.;  discharge  2128  cu.  ft.  per 
min.;  power  developed  80  H.P. 

Calculate  the  turbine  constants  including  the  characteristic 
speed. 


FIG.  205. 


FIG.  206.  FIG.  207. 

Solution. — In  this  case,  from  Article  36, 


60V/i     "    60V25 
k,  5.62 


V64.4 


=  0.7; 


248  ELEMENTS  OF  HYDRAULICS 

2128 


Qi  _  7.0933 
19 


/\ 
V12/ 


=  __5 

K/S       "  25^25  " 

135.  Two  types  of  turbine  runner,  A  and  5,  are  to  be  compared. 
From  tests  it  is  known  that  runner  A  will  develop  a  maximum 
of  2080  H.P.  at  500  r.p.m.  under  100  ft.  head,  and  runner  B 
will  develop  4590  H.P.  at  580  r.p.m.  under  150  ft.  head.  Deter- 
mine which  of  these  types  is  the  higher  speed. 


136.  Show  that  to  transform  the  characteristic  speed  N8 
from  the  English  to  the  metric  system  it  is  necessary  to  multiply 
by  the  coefficient  4.46;  that  is  to  say,  if  the  horse  power  and 
head  are  expressed  in  foot-pound  units,  and  N8  in  the  metric 
system,  we  have  the  relation 

A7  VlLR 

Ns  = 


137.  Five  two-runner  Francis  turbines  installed  in  the  power 
house  of  the  Pennsylvania  Water  and  Power  Co.  at  McCalPs 
Ferry   on  the   Susquehanna   River   are   rated   at    13,500   H.P. 
each  under  a  head  of  53  ft.  at  a  speed  of  94  r.p.m.     The  quantity 
of  water  required  per  turbine  is  2800  cu.  ft.  per  second.     Calculate 
from  this  rating  the  characteristic  speed,  efficiency,  and  other 
turbine  constants.  ' 

138.  Four  two-runner  Francis  turbines  operating  in  the  Little 
Falls  plant  of  the  Washington  Water  Power  Co.  have  a  nominal 
power  capacity  of  9000  H.P.  each  under  a  head   of  66  ft.  at  a 
speed  of  150  r.p.m.     The  quantity  of  water  required  per  turbine 
is  1500  cu.  ft.  per  second.     From  this  rating  calculate  the  charac- 
teristic speed,  efficiency,  and  specific  constants  for  these  units. 

139.  The  upper  curve  shown  in  Fig.   208   is  the  official  effi- 
ciency test  curve  of  the  9000  H.P.  turbines,  built  by  the  I.  P. 


HYDRODYNAMICS 


249 


Morris  Co.  for  the  Washington  Water  Power  Co.  These  wheels 
are  of  the  horizontal  shaft,  two-runner,  central  discharge  type, 
with  volute  casings.  Head  66  ft.,  speed  150  r.p.m.,  and  rated 
runner  diameter  6  ft.  2  in. 

The  lower  curve  shown  in  the  figure  is  derived  from  a  test  at 
Holyoke  of  a  homologous  experimental  runner  having  a  rated 
diameter  of  2  ft.  8-13/14  in.  These  curves  are  almost  identical 
in  shape,  the  efficiency  of  the  large  units  exceeding  by  a  small 
margin  that  of  the  experimental  runner. 


70 

GO 
t» 
§50 


10 


90 


50 


0   500  1000  1500  2000  2500  3000  3500  4000  4500  5000  5500  6000  6500  7000  7500  8000  3500  9000  9500 

Horse  Power 

FIG.  208. 

Calculate  the  discharge  and  characteristic  speed  at  maximum 
efficiency,  and  from  these  results  compute  the  specific  constants, 

140.  In  testing  a  hydraulic  turbine  it  was  found  by  measure- 
ment that  the  amount  of  water  entering  the  turbine  was  8000 
cu.  ft.  per  minute  with  a  net  fall  of  10.6  ft.     The  power  devel- 
oped was  measured  by  a  friction  brake  clamped  to  a  pulley. 
The  length  of  brake  arm  was  12  ft.,  reading  on  scales  400  lb.,  and 
speed  of  pulley  100  r.p.m.    Calculate  the  efficiency  of  the  turbine. 

141.  One  of  a  series  of  65  tests  of  a  31-in.  Wellman-Seaver- 
Morgan  turbine  runner  gave  the  following  data:1 

Gate  opening  75  per  cent.;  head  on  runner  17.25  ft.;  speed 
186.25  r.p.m.;  discharge  63.12  cu.  ft.  per  second;  power  developed 
111.66  H.P. 

Calculate  the  efficiency  and  the  various  turbine  constants. 

142.  One  of  a  series  of  82  tests  of  a  30-in.  Wellman-Seaver- 
Morgan  turbine  runner  gave  the  following  data:2 

1  Characteristics  of  Modern  Hyd.  Turbines,  C.  W.  Larner,  Trans.  Am. 
Soc.  C.  E.,  Vol.  LXVI  (1910),  pp.  306-386. 

2  Ibid. 


250 


ELEMENTS  OF  HYDRAULICS 


Gate  opening  80.8  per  cent.;  head  on  runner  17.19  ft.;  speed 
206  r.p.m.;  discharge  85.73  cu.  ft.  per  second;  power  developed 
146.05  H.P. 

Calculate  the  efficiency  and  the  other  turbine  constants. 

143.  Four  of  the  turbines  of  the  Toronto  Power  Co.  at  Niagara 
Falls  are  of  the  two-runner  Francis  type,  with  a  nominal  develop- 
ment of  13,000  H.P.  each  under  a  head  of  133  ft.  at  a  speed  of 
250  r.p.m.  The  quantity  of  water  required  per  turbine  is  1060 
cu.  ft.  per  second. 

Calculate  the  efficiency,  characteristic  speed  and  specific 
turbine  coefficients  for  these  units. 


90 


70 


CO 


SO 


10 


0   400  800  1200  1COO  2000  2400  2800  3200  3GOO  4000  4400  4800  5200  5600  GOOD  6400 
Horse  Power 

FIG.  209. 

144.  The  upper  curve  shown  in  Fig.  209  is  the  official  test 
curve  of  the  6000  H.P.  turbines  designed  by  the  I.  P.  Morris 
Co.  for  the  Appalachian  Power  Co.  The  rated  runner  diameter 
is  7  ft.  6-1/4  in.,  head  49  ft.,  and  speed  116  r.p.m.  These  turbines 
are  of  the  single  runner,  vertical  shaft  type. 

The  lower  curve  is  derived  from  a  test  at  Holyoke  of  the 
small,  homologous,  experimental  runner,  having  a  rated  diameter 
of  27-3/8  in.  The  curves  are  identical  in  shape,  but  owing  to  the 
better  arrangement  of  water  passages  in  the  large  plant,  its 
efficiency  considerably  exceeds  that  of  the  experimental  runner. 
It  may  also  be  noted  that  the  efficiency  shown  on  this  diagram 
is  the  highest  ever  recorded  in  a  well-authenticated  test. 

Calculate  the  discharge  and  characteristic  speed  at  maximum 
efficiency,  and"  from  these  results  compute  the  specific  turbine 
constants. 


HYDRODYNAMICS 


251 


145.  The  following  data,  taken  from  the  official  Holyoke 
test  reports,  give  the  results  of  tests  made  on  a  35-in.  vertical 
Samson  turbine  built  by  the  James  Leffel  Co.  of  Springfield, 
Ohio.  Calculate  the  turbine  constants  and  characteristic  speeds. 

TESTS  OF  35-iN.   VERTICAL  SAMSON  TURBINE 


.  Gate  opening 

Head  on 
wheel  in 
feet 

Speed 
in  rev. 
per  min. 

Discharge 
in.  cu.  ft. 
per  sec.    . 

Horse 
power 
developed 

Efficiency 
in  per 
cent. 

Full  gate  
0.9  gate  
0.8  gate  
0  75  gate.  . 

16.57 
16.69 
16.78 
16  86 

187 
191 
189 

187 

120.61 
114.35 
105.10 
100  29 

188.27 
188.88 
179.87 
172  57 

83.06 

87.26 
89.93 
89  99 

0.7  gate  
0.6  gate...... 
0.5  gate  

17.08 
17.23 
17.47 

188 
185 

188 

92.83 
77.15 
66.89 

160  .  03 
128.22 
108.72 

88.99 
85.05 
82.03 

146.  The  speed  and  water  consumption  of  a  turbine  vary  as 
the  square  root  of  the  head  ( V  h),  and  the  power  varies  as  the 
square  root  of  the  cube  of  the  head  ( V  h3) .     Thus  if  the  head  on 
a  wheel  is  multiplied  by  4,  the  speed  and  discharge  will  be 
multiplied  by  2  and  the  power  by  8. 

Given  that  a  12-in.  turbine  under  12-ft.  head  develops  14 
H.P.  at  480  r.p.m.  using  762  cu.  ft.  of  water  per  minute,  find  the 
power,  speed  and  discharge  for  the  same  turbine  under  48-ft. 
head. 

147.  On  page  252  is  given  a  rating  table  of  turbines  manu- 
factured by  the  S.  Morgan  Smith  Co.  of  York,  Pa.,  computed 
from  actual  tests  of  each  size  turbine  under  the  dynamometer  at 
the  Holyoke  testing  flume. 

Calculate  the  nominal  efficiency  and  characteristic  speed  for 
each  size  runner,  and  determine  whether  it  is  of  the  low,  medium 
or  high  speed  type. 

NOTE. — Data  of  this  kind  may  be  used  by  the  instructor  as 
problem  material  for  an  entire  class  without  duplicating  results, 
the  final  results  being  collected  and  tabulated,  thus  serving  as 
a  check  on  the  calculations  and  also  showing  the  range  of  the 
constants  involved. 

148.  On  pages  253,  254  and  255  is  given  a  rating  table  of 
Victor  Turbines  manufactured  by  the  Platt  Iron  Works  Co., 
Dayton,  Ohio. 

Calculate  the  nominal  efficiency,  characteristic  speed,  and 
speed  and  capacity  constants  for  each  diameter  and  head. 


252 


ELEMENTS  OF  HYDRAULICS 


HYDRODYNAMICS 


253 


all 


Hor 
cub 


.s       M 

fib! 

S3 
^    o 


TH  r^  co 

(M    OS    i—  l 

TH     »O     OS 


1>      TH 

TH     IO 


CO  CO  TH 
TH  CO  OS 
rH  »O  00 


TH 
TH    to 


o  co  t^. 

rH     10     00 


CO    to 

rH     IO 


CO   Ol   CO 

OS   00   TH 

TH    00 


CO 
OS 


O   t»   CO 

OS  TH   W 
TH    00 


CO 

oo 


O   CO 

co  TH 

TH    00 


oo 

00 


CO    Ofl    O1 

00   1>    (N 


l>    to   to 

t^     TH     rH 


CO    OS    O 

TH    CO    1> 

01 


TH     OS     CO 

CO   CO   OS 

rH     CO     CO 


IO    rH    10 

to   TH    00 

THCOCO 


CO 


to  oo  os 

TH    OO  CO 

rH     OQ  CO 

OQ 


1>    OS    r-  I 

O1    to    CO 
CO 


eo 

CO 


tO    CO  OS 

10    (N  OS 

O5    to  iO 

co 


b-      tO  TH 

TH     00  OS 

O*   TH  to 

CO 


O   00  t^ 

TH     TH  GO 

<N    TH  »o 

CO 


(N    O    OS 

CO       TH       b- 


TH 

(N 


<N    CO    to 

CO 


CO    TH  b- 

rH    CO  CO 

O}    CO  to 

CO 


OS    CO 
Oq    »0 

co 


O    »0  CO 

b-    OS  <M 

co  oo  to 
TH 


oo  co  oo 

IO     TH     rH 

co  oo  to 


00   CO   C^ 

Tt<     OS     rH 

CO    l>    lO 


CO  iO  i—  i 
<M  OS  O 
CO  <£>  10 


CO  O 
OS  OS 
10  TjH 


CO 


(N   1>    (N 

3§3 
CO 


0   O    »O 

OS     TtH     00 

<N    lO    rt< 


»O    00    OS 

oo  oo  t^ 


O    rH    TH 

TH    CO    CO 


i— I    TH     CO        I      OS 


TH    CO 

OS    OS 
CO    1> 


to   to 
N-    00 

CO   J> 


CO   TH 
to   t» 

CO   I> 


pe 


Cubi 
Revo 


tO 
0? 


eet 
ions 


Horse  power. 
Cubic 
Revolu 


<N    CO  TH 

o  to  10 

(M    (M  IO 

co 


i>  oo 

srH 

CO 


o  TH 

00    CO 


rH     O 

co 


powe 
feet 


o 
ub 


H 
C 
R 


1^    CO    l> 

(N    rJH    rjn 


>O 
<M 


«o  o 

CO     i-H 

(N   <N 


o 

IO 


C<J     10 
(N    rH 


rse  power 
bic  fee 


Ho 
Cu 
Re 


TH 

.OOTH 


10  oo 


CO    00 
t^    CO 


to  co  co 


OO 
TH 


CO     b- 


CO 
O 


q  01  os 

Ol    !>•    OS 

CO    iO    CO 


IO  TH  00 
OS  (N  00 
(M  TH  CO 


Horse  power 
Cubic  feet  .. 
Revolutions  . 


(N 


O   00   (M 

t^       TH       t^ 

TH    TH    CO 


TH    OS 

>O    (N    CO 


l>   t^    CO 

TH   CO 
TH    (N 


OQ    TH    to 

TH     TH     CO 

IS 


tO  01  TH 
O  to  to 
TH  O  CO 


Os   CO  00 

00   CO  TH 

CO    OS  CO 

CO 


CO   iO   TH 
b-   CO   TH 

co  oo  co 


Horse 
Cubic 
Revolu 


254 


ELEMENTS  OF  HYDRAULICS 


S 


po 


9?  +=  "d 

*Sg 

2  -3 

S  3  g 

w  §;3 


revolutio 
per  minu 


^  rH  iO 
to  OS  CO 
I>  OS  CO 


0  OS  00 

i—  i  oo  10 


CO  <N 

CO  00 

CO  *O 

OS 


T^  l>  CO 
^  l>  TjH 
CO  Tj<  CO 


CO  <N  <N 

(M  1>  T^ 

co  co  co 


rH  tO  OS 
O  CO  CO 
CO  (N  CO 


rH  CO  Tt< 

oo  to  co 


!>•  r-  i 

CO 
CO 


0 


O  O  00 


oo 


O  >O  <N 

(N  <N  (N 

iO  00  CO 
00 


iO  O5 

O  i-H 

i>  co 
00 


o  -^  10 

GO  OS  r-H 

rt<  >O  CO 

" 


rH  »O  O 

CO  l>-  i—  i 

Tt<  Tt<  CO 

00 


Horse  powe 
Cubic  feet 
Revolutions 


CO  O5  <N 

i—  1  00  CO 

o  o  co 


CO  (M 

Oi  co 


O5  »O  >O 
10  rj<  (M 

00  OO  CO 


^  00 

iO  CO 


O  OS  ^ 

00  CO  TH 
l>  Tt<  CO 


oo  TH 

<N  rH 

i>  d 


CO  O  "tf 
O  00  O 
1>*  CO  CO 


O5  1>  »O 

<N  r^  O5 

CO  CO  (M 


CO  TH 
CO  OS 
g  (N 


TH  O5  CO 

os 

CO 

o 


oo  os  oo 

IO  CO  (M 


Horse  powe 
Cubic  feet 
Revolutions 


CO  (N 
Os  O 


eo  os 

Tt<  OS 


OS  »O  OS 

OS  OS  (M 
CO 


1>  00  "tf 

OS  OS  OO 
00  TJH  (M 


l>  CO 

CO  -^ 

00  CO 

CO 


CO  J> 

CO  00 


00   l>   TH 

!>•  CO  b— 

l>  00  (N 


TH  to  (N 

OS  1>  CO 

CO  CO  (N 

(N 


to  00 

O  to 

(N  (M 
(N 


Horse  powe 
Cubic  feet 


CO  rt^  OS 
t^  00  l>- 

(M  00  (N 


CO  TH  t^ 

(M  I>  (M 

TH  CO 


8  CO  »O 
Tt<  t^ 


(N  OS 

TH   TH 

TH  CO 


OS  00  »O 

00  rH  CO 


CO  OS  (N 

to  co  co 


00 
IO 


(N  "*  CO 

00  l>  iO 

OS  TJH  (N 


I>  O  Tt< 

T*  OS  »O 

OS  (N  (M 


00  OS  <N 


rH  tO  (N 
rH  (N  Tj< 

00  iO  (N 


OS  to 

£8 
•if 


power 
feet 


Ho 
Cu 
Revolu 


o 
ub 


OS 


IO  CO  00 

co  oo  >o 


(M  00  »0 

OS  00  iO 

CO  rH  (M 


OS  CO  CO 

TfH  00  IO 

CO  OS  (M 

rH  00 


JO  ^  OS 

O  OO  ^T* 

CO  1>*  C^l 

rH  00 
rH 

co  CO~I>~ 

CO  1>  Tt^ 

(N  iO  (N 


(N 

oo  co  ^ 

"^  O^  CO 
»O  t^  (M 

rH  rH 
(N 

00  O  (N 

OS  CO  CO 

rH  rH 

<N 

^  CO  (M 

rH  rH 

(N 


l>  CO  CO 
OS  00  <N 
CO  O  (N 


O  rH  (N 

00  CO  Tt< 

rH  rH  (M 

TH  00 


Tt<  ^ 

tO   Tj<   (N 

CO  00  (M 


OS  O  OO 
CO  IO  CO 


l>-  rH 

CO  CO  (M 


§co  »o 
CO  CO 
1>  (M 


CO  »O 
(N  CO  (M 


os  co  co 

tO  rH  CO 

O  iO  <M 


(N  CO  00 

rH  O  rH 
(M  rH 


rH  OS  CO 
O  (N  (M 


O  CO 

rH  00  (M 


O  TjH  CO 

00  CO  (M 
OS  O  (M 


IO  rH 

N  OS  rH 

rH  tO 

rH  03 


rH  00  OS 

§8 
rH  OS 


CO  CO  rH 

OS  CO  (M 
CO 


O  O 


po 
fe 


Hor 
Cub 


ee 
ions 


Horse  power. 
Cubic  feet 
Revolution 


HYDRODYNAMICS 


255 


O  ^  "tf 
CO  t^  <N 

OS  iO  <M 


GO  CO 
(M 


iO  CO  O 
TH  10  (M 

GO  O  (M 


OS  Tt<  GO 
to  OS  TH 


CO  OS  r^ 
O  <N  TH 

l>  to  (N 


CO  CO 
CO  iO 

T^  1—  I 


<N  *O  O 

GO  t^  TH 


(N  >O 

O5  O 

<M  (M 

GO 


OS  O5 

TH   1> 

(N 


s 


co  co  os 

CO  O5  OS 
GO  CO  TH 


O  GO  OS 

GO  O  TH 


§^  (M 

CO  OS 

TH  CO  '" 

<M 


s-ss 

CO   TH   T—t 

TH  CO 


TH  CO  CO 
CO  <N  GO 
to  GO  TH 


co  Tt^  co 

O  O  GO 

»O  »O  TH 

rH  IO 


(N  I>  CO 

co  os  t^ 

CO  ^  rH 


CO  rH 

(N 

co 


CO  t^  CO 
1>  •<*!  OS 

CO  0  rH 

<N  (N 

CO 


00  TtH 

TH  OS 


(M   TH 

CO 


CO  GO  OS 
(N  CO  TH 


OS  GO  CO 

TH  CO  GO 

O  CO  TH 

<N  0 


CO  OS  TH 

00  CO  GO 


CO  iO  OS 

TH   TH   !>. 


(N 

O  OS  00 


•*  (M  CO 

GO  OS  1> 

CO  »O  TH 

TH  GO 


O  t^  •* 
<N  O  l> 

CO  (N  rH 


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256 


ELEMENTS  OF  HYDRAULICS 


149.  Fig.  210  shows  a  vertical  section  of  the  10,800-H.P. 
turbines  designed  by  the  I.  P.  Morris  Co.  for  the  Cedar 
Rapids  Mfg.  and  Power  Co.  The  rated  diameter  of  these  tur- 
bines is  11  ft.  10-1/2  in.,  head  30  ft.,  and  speed  55.6  r.p.m. 

These  turbines  are  at  present  the  largest  in  the  world,  and  it 
may  be  noted  that  all  the  latest  features  have  been  incorporated 
in  the  design,  namely,  volute  casings  and  draft  tubes  molded  in 


FIG.  210. 

the  concrete;  cast-iron  speed  rings  supporting  the  concrete,  genera- 
tor and  thrust  bearing  loads  from  above;  lignum  vitse  turbine 
guide  bearing;  thrust  bearing  support  located  above  the  generator ; 
Kingsbury  thrust  bearing  with  roller  auxiliary;  and  pneumatic 
brakes  acting  on  the  rotor  of  the  generator. 

Calculate  the  characteristic  speed  from  the  rating  given  above, 
and  from  the  table  on  page  188  determine  to  which  speed  type 
it  belongs. 

150.  The  following  table  gives  the  results  of  20  tests  out  of 
a  total  of  66  made  Sept.  3-4,  1912,  at  the  testing  flume  of  the 


HYDRODYNAMICS 


257 


Holyoke  Water  Power  Co.  on  a  24-in.  Morris  turbine  type  "O" 
runner.  Calculate  the  characteristic  speed  for  each  test,  and 
note  its  high  value  in  test  number  10. 

NOTE. — Two  of  the  wheels  for  the  Keokuk  installation  and 
nine  wheels  for  the  Cedar  Rapids  plant  are  built  with  this  type 
of  runner,  the  large  wheels  being  geometrically  similar  to  the 
experimental  wheel  tested  at  Holyoke. 

REPORT  OF  TESTS  OF  A24-IN.  MORRIS  TURBINE,  TYPE  "O"  RUNNER  MADE 
IN  THE  TESTING  FLUME  OF  THE  HOLYOKE  WATER  POWER  Co. 


Number 
of  experi- 
ment 

Open- 
ing of 
speed 
gate  in 
inches 

Per  cent. 
of  full 
dis- 
charge of 
wheel 

Head 
on 
wheel  in 
feet 

Speed  in 
rev.     per 
min. 

Dis- 
charge 
in  cu. 
ft.  per 
sec. 

Horse 
power 
devel- 
oped 

Effi- 
ciency 
in  per 
cent. 

1 

3.0 

0.792 

17.39 

344.25 

84.66 

103.98 

62.31 

2 

3.0 

0.779 

17.40 

298.00 

83.35 

126.01 

76.66 

3 

3.0 

0.775 

17.39 

275.75 

82.91 

133.  2& 

81.55 

4 

3.0 

0.778 

17.36 

257.50 

83.13 

139.99 

85.59 

5 

3.0 

0.779 

17.33 

249.20 

83.13 

143.01 

87.58 

6 

3.0 

0.782 

17.30 

241.25 

83.35 

145.73 

89.17 

7 

3.0 

0.779 

17.30 

235.50 

83.13 

146.53 

89.89 

8 

3.0 

0.781 

17.28 

238.25 

83.24 

146  .  80 

90.05 

9 

3.0 

0.783 

17.27 

240.20 

83.46 

146.55 

89.71 

10 

3.0 

0.781 

17.23 

236.80 

83.13 

146.62 

90.32 

11 

3.0 

0.762 

17.31 

214.00 

81.28 

138.97 

87.15 

12 

3.0 

0.782 

17.24 

237.20 

83.24 

146.51 

90.08 

13 

3.0 

0.744 

17.34 

185.50 

79.44 

128.86 

82.54 

14 

3.5 

0.902 

17.15 

357.20 

95.73 

107.89 

57.98 

15 

3.5 

0.900 

17.10 

320.50 

95.38 

135.52 

73.31 

16 

3.5 

0.903 

17.06 

287.75 

95.61 

156.44 

84.62 

17 

3.5 

0.892 

17.14 

265.75 

94.70 

160.53 

87.26 

18 

3.5 

0.871 

17.17 

243.25 

92.55 

154.29 

85.67 

19 

3.5 

0.903 

17.15 

280.20 

95.84 

160.80 

86.32 

20 

3.5 

0.904 

17.19 

278.60 

96.07 

164.09 

87.67 

17 


HYDRAULIC  DATA  AND  TABLES 


260 


ELEMENTS  OF  HYDRAULICS 


TABLE  1. — PROPERTIES  OF  WATER 
Density  and  Volume  of  Water 


Temp,  in 
degrees 
Centigrade 

Density 

Volume  of 
1  gram  in 
cu.  cm. 

Temp,  in 
degrees 
Centigrade 

Density 

Volume  of 
1  gram  in 
cu.  cm. 

0 

0.999874 

1.00013 

24 

0.997349 

.00266 

1 

0.999930 

1.00007 

26 

0  .  996837 

.00317 

2 

0.999970 

1  .  00003 

28 

0.996288 

.00373 

2 

0.999993 

1.00001 

30 

0.995705 

.00381 

4 

1.000000 

1.00000 

32 

0.995087 

.00394 

5 

0.999992 

1.00001 

35 

0.995098 

.00394 

6 

0.999970 

1  .  00003 

40 

0.99233 

.00773 

7 

0.999932 

1.00007 

45 

0.99035 

1.00974 

8 

0.999881 

1.00012 

50 

0.98813 

1.01201 

9 

0.999815 

1.00018 

55 

0.98579 

1.01442 

10 

0.999736 

1  .  00026 

60 

0.98331 

1.01697 

11 

0.999643 

1.00036 

65 

0.98067 

1.01971 

12 

0.999537 

1.00046 

70 

0.97790 

1.02260 

13 

0.999418 

1  .  00058 

75 

0.97495 

1.02569 

14 

0.999287 

1.00071 

80 

0.97191 

1.02890 

16 

0.998988 

1.00101 

85 

0.96876 

1.03224 

18 

0.998642 

1.00136 

90 

0.96550 

1.03574 

20 

0.998252 

1.00175 

95 

0.96212 

1.03938 

22 

0.997821 

1.00218 

100 

0.95863 

1.04315 

Weight  of   Water 


Temp,  in 
degrees 
Fahrenheit 

Weight  in 
pounds  per 
cu.  ft. 

Temp,  in 
degrees    • 
Fahrenheit 

Weight  in 
pounds  per 
cu.  ft. 

Temp,  in 
degrees 
Fahrenheit 

Weight  in 
pounds  per 
cu.  ft. 

32 
40 
50 
60 
70 
80 
90 

62.42 
62.42 
62.41 
62.37 
62.31 
62.23 
62.13 

100 
110 
120 
130 
140 
150 
160 

62.02 
61.89 
61.74 
61.56 
61.37 
61.18 
60.98 

170 
180 
190 
200 
210 
212 

60.77 
60.55 
60.32 
60.07 
59.82 
59.56 

TABLES 


261 


TABLE  2. — HEAD  AND  PRESSURE  EQUIVALENTS 
Head  of  Water  in  Feet  and  Equivalent  Pressure  in  Pounds  per  Sq.  In. 


Feet 
head 

Pounds  per 
sq.  in. 

Feet 
head 

Pounds  per 
sq.  in. 

Feet 
head 

Pounds  per 
sq.  in. 

1 

0.43 

55 

23.82 

190 

82.29 

2 

0.87 

60 

25.99 

200 

86.62 

3 

1.30 

65 

28.15 

225 

97.45 

4 

1.73 

70 

30.32 

250 

108.27 

5 

2.17 

75 

32.48 

275 

119.10 

6 

2.60 

80 

34.65 

300 

129.93 

7 

3.03 

85 

36.81 

325 

140.75 

8 

3.40 

90 

38.98 

350 

151.58 

9 

3.90 

95 

41.14 

375 

162.41 

10 

4.33 

100 

43.31 

400 

173.24 

15 

6.50 

110 

47.64 

500 

216.55 

20 

8.66 

120 

51.97 

600 

259.85 

25 

10.83 

130 

56.30 

700 

303  .  16 

30 

12.99 

140 

60.63 

800 

346.47 

35 

15.16 

150 

64.96 

900 

389.78 

40 

17.32 

160 

69.29 

1000 

433.09 

45 

19.49 

170 

73.63 

50 

21.65 

180 

77.96 

Pressure  in  Pounds  per  Sq.  In.  and  Equivalent  Head  of  Water  in  Feet 


Pounds  per 
sq.  in. 

Feet 
head 

Pounds  per 
sq.  in. 

Feet 
head 

Pounds  per 
sq.  in. 

Feet 
head 

1 

2.31 

55 

126.99 

180 

415.61 

2 

4.62 

60 

138  .  54 

190 

438.90 

3 

6.93 

65 

150.08 

200 

461  .  78 

4 

9.24 

70 

161.63 

225 

519.51 

5 

11.54 

75 

173.17 

250 

577.24 

6 

13.85 

80 

184.72 

275 

643.03 

7 

16.16 

85 

196.26 

300 

692  .  69 

8 

18.47 

90 

207.81 

325 

750.41 

9 

20.78 

95 

219.35 

350 

808.13 

10 

23.09 

100 

230.90 

375 

865.89 

15 

34.63 

110 

253.98 

400 

922.58 

20 

46.18 

120 

277.07 

500 

1154.48 

25 

57.72 

125 

288  .  62 

30 

69.27 

130 

300  .  16 

35 

80.81 

140 

323.25 

40 

92.36 

150 

346  .  34 

45 

103  .  90 

160 

369  .  43 

50 

115.45 

170 

392.52 

262 


ELEMENTS  OF  HYDRAULICS 


TABLE  3. — DISCHARGE  EQUIVALENTS 


Gallons 
per 

inin. 

Cubic 
feet  per 
sec. 

Cubic  feet 
per 
min. 

Gallons 
per 
hour 

Gallons 
per  24 
hours 

Bbls.  per 
minute, 
42  gal. 
bbl. 

Bbls.  per 
hour,  42 
gal.  bbl. 

Bbls.  per 
24  hours,  42 
gal.  bbl. 

10 

1  3368 

600 

14,400 

0.24 

14  28 

342  8 

12 

1  .  6042 

720 

17,280 

0.29 

17.  14 

411  4 

15 

2.0052 

900 

21,600 

0.36 

21.43 

514  3 

18 

2  4063 

1,080 

25,920 

0.43 

25.71 

617   1 

20 

2  6733 

1,200 

28,800 

0.48 

28.57 

685  7 

25 

3.342 

1,500 

36,000 

0.59 

35.71 

857  0 

27 

3.609 

1,620 

38,880 

0.64 

38.57 

925  0 

30 

4.001 

1,800 

43,200 

0.71 

42.85 

1,028  0 

35 

4  678 

2  100 

50,400 

0  83 

50.0 

1  200  0 

36 

4  812 

2  160 

51,840 

0  86 

51  43 

1  234  0 

40 
45 
50 
60 
70 

0.1 

5.348 
6.015 
6.684 
8.021 
9.357 

2,400 
2,700 
3,000 
3,600 
4,200 

57,600 
64,800 
72,000 
86,400 
100,800 

0.95 
1.07 
1.19 
1.43 
1.66 

57.14 
64.28 
71.43 
85.71 
100.0 

1,371.0 
1,543.0 
1,714.0 
2,057.0 
2  400  0 

75 

10  026 

4,500 

108,000 

1  78 

107  14 

2  570  0 

80 

10  694 

4,800 

115,200 

1  90 

114  28 

2  742  0 

90 
100 

0.2 

12.031 
13  368 

5,400 
6,000 

129,600 
144,000 

2.14 
2  39 

128.5 
142  8 

3,085.0 
3  428  0 

125 

16.710 

7,500 

180,000 

2  98 

178.6 

4  286  0 

135 
150 

0.3 

18.046 
20  052 

8,100 
9  000 

194,400 
216  000 

3.21 
3  57 

192.8 
214  3 

4,628.0 
5  143  0 

175 

23  394- 

10  500 

252  000 

4  16 

250  0 

6  000  0 

180 
200 

0.4 

24.062 
26  736 

10,800 
12  000 

259,200 
288,000 

4.28 
4  76 

257.0 

285  7 

6,171.0 
6  857  0 

225 
250 

0.5 

30.079 
33.421 

13,500 
15,000 

324,000 
360,000 

5.35 
5  95 

321.4 
357.1 

7,714.0 
8  570  0 

270 
300 
315 
360 
400 

0.6 
0.8 

36.093 
40.104 
42.109 
48.125 
53  472 

16,200 
18,000 
18,900 
21,600 
24  000 

388,800 
432,000 
453,600 
518,400 
576,000 

6.43 
7.14 
7.5 
8.57 
9  52 

385.7 
428.5 
450.0 
514.3 

571  8 

9,257.0 
10,284.0 
10,800.0 
12,342.0 
13  723*0 

450 
500 

1.0 

60.158 
66  842 

27,000 
30  000 

648,000 
720,000 

10.7 
11  9 

642.8 
714  3 

15,428.0 
17  143  0 

540 
600 

1.2 

72.186 
80.208 

32,400 
36,000 

777,600 
864,000 

12.8 
14.3 

771.3 
857.1 

18,512.0 
20,570  0 

630 
675 
720 
800 
900 
,000 

1.4 
1.5 
1.6 

2.0 

84.218 
90.234 
96.25 
106.94 
120.31 
133.68 

37,800 
40,500 
43,200 
48,000 
54,000 
60,000 

907,200 
972,000 
1,036,800 
1,152,000 
1,296,000 
1,440,000 

15.0 
16.0 
17.0 
19.05 
21.43 
23  8 

900.0 
964.0 
1,028.0 
1,142.0 
1,285.0 
1,428  0 

21,600.0 
23,143.0 
24,685.0 
27,387.0 
30,857.0 
34  284  0 

,125 
,200 
,350 
,500 

2.5 
3.0 

150.39 
160.42 
180.46 
200.52 

67,500 
72,000 
81,000 
90,000 

1,620,000 
1,728,000 
1,944,000 
2,160,000 

26.78 
28.57 
32.14 
35  71 

1,607.0 
1,714.0 
1,928.0 
2  142  0 

38,571.0 
41,143.0 
46,085.0 
51  427  0 

,575 
,800 
2,000 

3.5 
4.0 

210.54 
240.62 
267  36 

94,500 
108,000 
120  000 

2,268,000 
2,592,000 
2  880  000 

37.5 

42.85 
47  64 

2,250.0 
2,571.0 
2  857  0 

54,000.0 
61,710.0 
68  568  0 

2,025 
2,250 

4.5 

270.70 
300  78 

121,500 
135  000 

2,916,000 
3  240  000 

48.21 
53  57 

2,892.0 
3  214  0 

69,425.0 
77  143  0 

2,500 

334  21 

150  000 

3  600  000 

59  52 

3  571  0 

85  704  0 

2,700 
3,000 

6.0 

360.93 
401.04 

162,000 
180,000 

3,880,000 
4,320,000 

64.3 
71.43 

3,857.0 
4,285.0 

92,572.0 
102,840.0 

TABLES 


263 


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264 


ELEMENTS  OF  HYDRAULICS 


TABLE  5. — SPECIFIC  WEIGHTS  OP  VARIOUS  SUBSTANCES 


Air,  press.  76  cm.  Hg., 
0°  C  
Alcohol 

0.001293 
0.79 

Iron,  cast  
pure  
steel  . 

7.03-  7.73 

7.85-  7.88 
7  60-  7  80 

Aluminium,  pure  

2.583 

wrought  

7.79-  7.85 

commercial  
Basalt 

2.7  -  2.8 
2.4  -  3.3 

Lead  
Lime  mortar 

11.21-11.45 
16-18 

Bismuth  
Brass  

9.76-  9.93 

7.8  -  8.7 

Limestone  
Magnesium  .  . 

2.4  -  2.8 
1.69-  1  75 

Brick 

1.4  -  2  3 

Marble  . 

25-29 

Cadmium  

8.54-  8.69 

Mercury  at  0°  C  

13.596 

Carbon,  charcoal  
diamond 

1.45-  1.70 
3.49-  3  53 

Nickel  
Oil 

8.57-  8.93 
0  91-  0  94 

graphite  
Coal,  hard 

2.17-  2.32 
1.2  -  1.8 

Platinum,  cast  
wire  and  foil 

21.48-21.50 
21.2  -21  7 

Copper,  cast  

8.3  -  8.92 

Quartz  

2.3  -  2.7 

electrolytic  
wire  

8.88-  8.95 
8.93-  8.95 

Rubber  
Sand. 

0.93 
1.2  -  1.9 

Cork 

0.24 

Sandstone 

19-27 

Earth  

1.4  -  2.8 

Seawater  

1.02-  1.03 

Gold  

19.30-19.34 

Silver  

10.42-10  57 

Glass 

2  5  -  3  8 

Timber,  oak 

0  62-  1  17 

Granite  

2.5  -  3.0 

fir  

0.5  -  0.9 

Hydrogen,  press.  76  cm. 

poplar 

0.35-  1  02 

He.,  0°  C 

0  0000894 

Tin 

6  97-  7  37 

Ice  

0.926 

Zinc  

6.86-  7.24 

TABLES 


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266 


ELEMENTS  OF  HYDRAULICS 


TABLE  7. — CAPACITY  OF  RECIPROCATING  PUMPS 

Capacity,  or  piston  displacement,  of  reciprocating  pumps  in  gallons  per  single  stroke 


Diameter 
of 
cylinder, 
inches 

Length  of  stroke  in  inches 

2 

3 

-4 

5 

6 

7 

8 

9 

10 

1-1/4 

0.0106 

0.0159 

0.0212 

0.0266 

0.0319 

0.0372 

0.0425 

0.0478 

0.0531 

1-3/8 

0.0128 

0.0192 

0.0256 

0.0321 

0.0385 

0.0419 

0.0513 

0.0578 

0.0642 

1-1/2 

0.0153 

0.0229 

0.0306 

0.0382 

0  .  0459 

0.0535 

0.0612 

0.0688 

0.0765 

1-3/4 

0  .  0208 

0.0312 

0.0416 

0.0521 

0.0625 

0.0729 

0.0833 

0.0937 

0.1041 

2 

0.0272 

0  .  0408 

0.0544 

0.068 

0.0816 

0.0952 

0.1088 

0.1224 

0.136 

2-1/4 

0  .  0344 

0.0516 

0.0688 

0.086 

0.1033 

0.1205 

0.1377 

0.1548 

0.1721 

2-1/2 

0  .  0425 

0.0637 

0.0850 

0.1062 

0.1275 

0.1487 

0.17 

0.1912 

0.2125 

2-3/4 

0.0514 

0.0771 

0.1028 

0.1285 

0.1543 

0.1799 

0.2057 

0.2313 

0.2571 

3 

0.0612 

0.0918 

0.1224 

0.1530 

0.1836 

0.2142 

0.2448 

0.2754 

0.306 

3-1/4 

0.0718 

0.1077 

0.1436 

0.1795 

0.2154 

0.2513 

0.2872 

0.3231 

0.3594 

3-1/2 

0.0833 

.0.1249 

0.1666 

0.2082 

0.2499 

0.2915 

0.3332 

0.3748 

0.4165 

3-3/4 

0  .  0956 

0.1434 

0.1912 

0.239 

0.2868 

0.3346 

0.3824 

0.4302 

0.478 

4 

0.1088 

0.1632 

0.2176 

0.272 

0.3264 

0.3808 

0.4352 

0.4896 

0.544 

4-1/4 

0.1228 

0.1842 

0.2456 

0.307 

0.3684 

0.4298 

0.4912 

0.5526 

0.6141 

4-1/2 

0.1377 

0.2065 

0.2754 

0.3442 

0.4131 

0.4819 

0.5508 

0.6196 

0.6885 

4-3/4 

0.1534 

0.2301 

0.3068 

0.3835 

0.4602 

0  .  5369 

0.6136 

0  .  6903 

0.7671 

5 

0.17 

0.2550 

0.34 

0.425 

0.51 

0.595 

0.68 

0.765 

0.85 

5-1/4 

0.1874 

0.2811 

0.3748 

0.4685 

0  .  5622 

0.6559 

0.7496 

0.8433 

0.9371 

5-1/2 

0.2057 

0.3085 

0.4114 

0.5,142 

0.6171 

0.7199 

0.8228 

0.9256 

1  .  0285 

5-3/4 

0.2248 

0.3372 

0.4496 

0.562 

0.6744 

0.7868 

0.8992 

1.011 

1.124 

6 

0.2448 

0.3672 

0.4896 

0.612 

0.7344 

0.8568 

0.9792 

1.1016 

1  .  2240 

6-1/4 

0.2656 

0.3984 

0.5312 

0.6640 

0.7968 

0.9296 

.062 

1,195 

1.328 

6-1/2 

0.2872 

0.4308 

0.5744 

0.7182 

0.8610 

.0052 

.1488 

1.2926 

1.4364 

6-3/4 

0.3098 

0.4647 

0.6196 

0.7745 

0.9294 

.084 

.239 

1.394 

1.549 

7 

0.3332 

0.4998 

0.6664 

0.833 

0.9996 

.1662 

.3328 

1.4994 

1.666 

7-3/4 

0.4084 

0.6126 

0.8168 

1.021 

1.225 

.429 

.633 

1.837 

2.042 

8 

0.4352 

0.6528 

0.8704 

1.088 

1.3056 

.5232 

.7408 

1  .  9584 

2.176 

9 

0.5508 

0.8262 

1.1010 

1.377 

1  .  6524 

.9278 

2.2032 

2.4786 

2.754 

10 

0.68 

1.02 

1.36 

1.7 

2.04 

2.38 

2.72 

3.06 

3.4 

11 

0.8227 

1.2341 

1.6451 

2.057 

2.464 

2.879 

3.2911 

3.7258 

4.1139 

12 

0.9792 

1.468 

1.9584 

2.448 

2.9376 

3.4222 

3.9168 

4.4064 

4.896 

13 

1.149 

1.723 

2.297 

2.872 

3.445 

4.022 

4.596 

5.170 

5.745 

14 

1.332 

1.998 

2.665 

3.331 

3.997 

4.664 

5.33 

5.996 

6.663 

15 

1.529 

2.294 

3.059 

3.824 

4.589 

5.354 

6.119 

6.884 

7.649 

16 

1.74 

2.61 

3.48 

4.35 

5.22 

6.09 

6.96 

7.83 

8.703 

18 

2.202 

3.303 

4.404 

5.505 

6.606 

7.707 

8.808 

9.909 

11.01 

20 

2.720 

4.08 

5.440 

6.8 

8.16 

9.52 

10.88 

12.24 

13.6 

TABLES 


267 


TABLE  7. — CAPACITY  OF  RECIPROCATING  PUMPS — (Continued) 


Diameter 
of 
cylinder, 
inches 

Length  of  stroke  in  inches 

12 

14 

15 

16 

18 

20 

22 

•  24- 

1-1/4 

0.0637 

0.0743 

0.0797 

0.0848 

0  .  0955 

0.1062 

0.1168 

0.1274 

1-3/8 

0.077 

0.089 

0.0963 

0.1027 

0.1156 

0.12SO 

0.1408 

0.1541 

1-1/2 

0.0918 

0.1071 

0.1147 

0.1224 

0.1377 

0.1530 

0.1683 

0.1836 

1-3/4 

0.1249 

0.1457 

0.1562 

0.1666 

0.1874 

0.2082 

0.2290 

0  .  2499 

2 

0.1632 

0.1904 

0.204 

0.2176 

0.2448 

0.2720 

0.2992 

0.3264 

2-1/4 

0.2063 

0.241 

0.258 

0.2754 

0.3096 

0.344 

0.3784 

0.4128 

2-1/2 

0.255 

0.2975 

0.3187 

0.34 

0.3825 

0.4252 

0.4677 

0.51 

2-3/4 

0.3085 

0.3598 

0  .  3855 

0.4114 

0.4626 

0.5142 

0  .  5656 

0.617 

3 

0.3672 

0.4284 

0.459 

0.4896 

0.5508 

0.612 

0.6732 

0.7344 

3-1/4 

0.4312 

0.503 

0.5385 

0.5748 

0.6466 

0.7182 

0.79 

0.8624 

3-1/2 

0  .  4998 

0.5831 

0.6247 

0.6664 

0.7497 

0.833 

0.9163 

0.9996 

3-3/4 

0.5736 

0.6692 

0.687 

0.7648 

0.8605 

0.9561 

1.0517 

1.147 

4 

0.6528 

0.7616 

0.816 

0  .  8904 

0.9792 

1.088 

1  .  1968 

1  .  3056 

4-1/4 

0.7368 

0.8596 

0.921 

0.9824 

1.105 

1.228 

1  .  3508 

1.473 

4-1/2 

0.8262 

0.9639 

1.0327 

1.1016 

1  .  2393 

1.377 

1.5147 

1  .  6524 

4-3/4 

0.9204 

1.073 

1.15 

1.227 

1.380 

1.534 

1  .  6874 

1.84 

5 

1.02 

1.19 

1.275 

1.36 

1.53 

1.7 

1.87 

2.04 

5-1/4 

1.124 

1.311 

1.405 

1.499 

1.686 

1.874 

2.0614 

2.248 

5-1/2 

1.2342 

1.4399 

1  .  5427 

1  .  6456 

1.8513 

2.057 

2.2627 

2.4684 

5-3/4 

1.348 

1.573 

1.686 

1.789 

2.022 

2.248 

2.4728 

2.696 

6 

1.4688 

1.7136 

1  .  8362 

1.9584 

2.2032 

2.448 

2.6928 

2.9376 

6-1/4 

1.593 

1.859 

1.992 

2.124 

2.39 

2.656 

2.9216 

3.186 

6-1/2 

1.7955 

2.0109 

2.1546 

2.2982 

2.5885 

2.8728 

3.16 

3.4473 

6-3/4 

1.858 

2.168 

2.323 

2.479 

2.788 

3.098 

3.4078 

.3.716 

7 

1.9992 

2.3324 

2.499 

2.6656 

2.9988 

3.332 

3.6652 

3  .  9984 

7-3/4 

2.45 

2.858 

3.063 

3.266 

3.674 

4.084 

4.4924 

4.9 

8 

2.6112 

3.0464 

3.264 

3.4816 

3.9168 

4.352 

4.7872 

5  .  2224 

9 

3.3048 

3.8556 

4.131 

4.4064 

5.0572 

5.508 

6.0588 

6.6096 

10 

4.08 

4.76 

5.1 

5.44 

6.12 

6.8 

7.48 

8.16 

11 

4.9367 

5.7595 

6.1709 

6.5823 

7.4051 

8.2279 

9  .  0506 

9.8735 

12 

5.8752 

6.8544 

7.344 

7.833 

8.8128 

9.792 

10.7712 

11.7504 

13 

6.894 

8.042 

8.616 

9.192 

10.34 

11.49 

12.639 

13.78 

14 

7.994 

9.328 

9.993 

10;66 

11.99 

13.32 

14.652 

15.98 

15 

9.178 

10.70 

11.47 

12*23 

13.76 

15.29 

16.819 

18.35 

16 

10.44 

12.18 

13.05 

13.92 

15.66 

17.40 

19.14 

20.88 

18 

13.21 

15.41 

16.51 

17.61 

19.81 

22.02 

24.22 

26.42 

20 

16.32 

19.04 

20.4 

21.76 

24.48 

27.2 

29.92 

32.6 

268 


ELEMENTS  OF  HYDRAULICS 


TABLE  8. — CIRCUMFERENCES  AND  AREAS  OF  CIRCLES 

Diameters,    1/16  in.    up  to  and   including    120   in.     Advancing,    1/16   to    1;    1/8  to   50; 
1/4  to  80,  and  1/2  to  120 


Diam- 
eter, 
inches 

Circum- 
ference, 
inches 

Area, 
square 
inches 

Diame- 
ter, 
inches 

Circum- 
ference, 
inches 

Area, 
square 
inches 

Diame- 
ter, 
inches 

Circum- 
ference, 
inches 

Area, 
square 
inches 

1/16 

0.19635 

0.00307 

4-1/2 

14.137 

15.904 

9-5/8 

30  .  237 

72.759 

1/8 

0.3927 

0.01227 

4-5/8 

14.529 

16.800 

9-3/4 

30.630 

74  .  662 

3/16 

0.5890 

0.02761 

4-3/4 

14.922 

17.720 

9-7/8 

31.023 

76.588 

1/4 

0.7854 

0.04909 

4-7/8 

15.315 

18.665 

5/16 

0.9817 

0.07670 

10 

31.416 

78  .  540 

3/8 

1.1781 

0.1104 

5 

15.708 

19.635 

10-1/8 

31.808 

80.515 

7/16 

1  .  3744 

0.1503 

5-1/8 

16.100 

-20.629 

10-1/4 

32.201 

82.516 

1/2 

1  .  5708 

0.1963 

5-1/4 

16.493 

21  .  647 

10-3/8 

32.594 

84.540 

9/16 

1  .  7771 

0.2485 

5-3/8 

16.886 

22.690 

10-1/2 

32.986 

86.590 

5/8 

1  .  9635 

0  .  3068 

5-1/2 

17.278 

23  .  758 

10-5/8 

33.379 

88  .  664 

11/16 

2.1598 

0.3712 

5-5/8 

17.671 

24.850 

10-3/4 

33.772 

90.762 

3/4 

2.3562 

0.4417 

5-3/4 

18.064 

25.967 

10-7/8 

34.164 

92.885 

13/16 

7/8 

2.5525 
2  .  7489 

0.5185 
0.6013 

5-7/8 

18.457 

27.108 

11 

34  .  558 

95.033 

15/16 

2.9452 

0  .  6903 

6 

18.849 

28.274 

11-1/8 
11-1/4 

34  .  950 
35.343 

97.205 
99  .  402 

1 
1-1/8 
1-1/4 
1-3/8 
1-1/2 

3.1416 
3.5343 
3.9270 
4.3197 
4.7124 

0.7854 
0.9940 
1.2271 
1.4848 
1.7671 

6-1/8 
6-1/4 
6-3/8 
6-1/2 
6-5/8 
6-3/4 

19.242 
19.635 
20.027 
20.420 
20.813 
21.205 

29.464 
30.679 
31.919 
33.183 
34.471 
35.784 

11-3/8 
11-1/2 
11-5/8 
11-3/4 

11-7/8 

35.735 
36.128 
36.521 
36.913 
37.306 

101.623 
103  .  869 
106.139 
108.434 
110.753 

1-5/8 
1-3/4 
1-7/8 

5.1051 
5.4978 
5.8905 

2  .  0739 
2.4052 
2.7621 

6-7/8 

7 

21  .  598 
21.991 

37.122 

38.484 

12    > 

12-1/8 
12-1/4 

37.699 
38.091 
38.484 

113.097 
115.466 
117.859 

2 

6  .  2832 

3.1416 

7-1/8 

22  .  383 

39.871 

12-3/8 

38  .  877 

120.276 

2-1/8 

6.6759 

3  .  5465 

7-1/4 

22.776 

41.282 

12-1/2 

39  .  270 

122.718 

2-1/4 

7.0686 

3.9760 

7-3/8 

23.169 

42.718 

12-5/8 

39  .  662 

125.184 

2-3/8 

7.4613 

4  .'4302 

7-1/2 

23.562 

44.178 

12-3/4 

40  .  055 

127.676 

2-1/2 

7.8540 

4.9087 

7-5/8 

23.954 

45.663 

12-7/8 

40.448 

130.192 

2-5/8 
2-3/4 

8.2467 
8.6394 

5.4119 
5.9395 

7-3/4 
7-7/8 

24.347 
24.740 

47.173 
48.707 

13 
13-1/8 

40.840 
41.233 

132.732 
135.297 

2-7/8 

9.0321 

6.4918 

13-1/4 

41.626 

137.886 

8 

25.132 

50.265 

13-3/8 

42.018 

140.500 

3 

9.4248 

7.0686 

8-1/8 

25.515 

51.848 

13-1/2 

42.411 

143.139 

3-1/8 

9.8175 

7.6699 

8-1/4 

25.918 

53  .  456 

13-5/8 

42.804 

145.802 

3-1/4 

10.210 

8.2957 

8-3/8 

26.310 

55.088 

13-3/4 

43.197 

148.489 

3-3/8 

10.602 

8.9462 

8-1/2 

26.703 

56.745 

13-7/8 

43.589 

151.201 

3-1/2 

10.995 

9.6211 

8-5/8 

27.096 

58.426 

3-5/8 

11.388 

10.320 

8-3/4 

27.489 

60.132 

14 

43.982 

153.938 

3-3/4 

11.781 

11.044 

8-7/8 

27.881 

61.862 

14-1/8 

44.375 

156.699 

3-7/8 

12.173 

11.793 

14-1/4 

44  .  767 

159.485 

9 

28.274 

63.617 

14-3/8 

45.160 

162.295 

4 

12.566 

12.566 

9-1/8 

28.667 

65  .  396 

14-1/2 

45.553 

165.130 

4-1/8 

12.959 

13.364 

9-1/4 

29  .  059 

67  .  200 

14-5/8 

45.945 

167.989 

4-1/4 

13.351 

14.186 

9-3/8 

29.452 

69  .  029 

14-3/4 

46.338 

170.873 

4-3/8 

13.744 

15.033 

9-1/2 

29.845 

70.882 

14-7/8 

46.731 

173.782 

TABLES 


269 


TABLE  8. — CIRCUMFERENCES  AND  AREAS  OF  CIRCLES — (Continued) 


Diame- 
ter 
inches 

Circum- 
ference 
inches 

Area 

square 
inches 

Diame- 
ter 
inches 

Circum- 
ference 
inches 

Area 
square 
inches 

Diame- 
ter 
inches 

Circum- 
ference 
inches 

Area 
square 
inches 

15 

47.124 

176.715 

21 

65.973 

346.361 

27 

84.823 

572.556 

15-1/8 

47.516 

179.672 

21-1/8 

66.366 

350.497 

27-1/8 

85.215 

577.870 

15-1/4 

47.909 

182.654 

21-1/4 

66.759 

354.657 

27-1/4 

85.608 

583.208 

15-3/8 

48.302 

185.661 

21-3/8 

67.151 

358.841 

27-3/8 

86.001 

588.571 

15-1/2 

48.694 

188.692 

21-1/2 

67.544 

363.061 

27-1/2 

86.394 

593.958 

15-5/8 

49.087 

191.748 

21-5/8 

67.937 

367.284 

27-5/8 

86.786 

599  .  370 

15-3/4 

49.480 

194.828 

21-3/4 

68.329 

371.543 

27-3/4 

87.179 

604  .  807 

15-7/8 

49.872 

197.933 

21-7/8 

68.722 

375.826 

27-7/8 

87.572 

610.268 

16 

50.205 

201.062 

22 

69.115 

380  .  133 

28 

87.964 

615.753 

16-1/8 

50.658 

204.216 

22-1/8 

69  .  507 

384.465 

28-1/8 

88.357 

621  .  263 

16-1/4 

51.051 

207.394 

22-1/4 

69  .  900 

388.822 

28-1/4 

88.750 

626.798 

16-3/8 

51.443 

210.597 

22-3/8 

70.293 

393.203 

28-3/8 

89.142 

632.357 

16-1/2 

51.836 

213.825 

22-1/2 

70.686 

397.608 

28-1/2 

89  .  535 

637.941 

16-5/8 

52.229 

217.077 

22-5/8 

71.078 

402.038 

28-5/8 

89  .  928 

643  .  594 

16-3/4 

52.621 

220.353 

22-3/4 

71.471 

406.493 

28-3/4 

90.321 

649  .  182 

16-7/8 

53.014 

223.654 

22-7/8 

71  .  864 

410.972 

28-7/8 

90.713 

654.837 

17 

53.407 

226.980 

23 

72.256 

415.476 

29 

91.106 

660.521 

17-1/8 

53.799 

230.330 

23-1/8 

72.649 

420.004 

29-1/8 

91.499 

666.277 

17-1/4 

54.192 

233.705 

23-1/4 

73.042 

424.557 

29-1/4 

91.891 

671.958 

17-3/8 

54  .  585 

237.104 

23-3/8 

73.434 

429.135 

29-3/8 

92.284 

677.714 

17-1/2 

54  .  978 

240.528 

23-1/2 

73.827 

433.731 

29-1/2 

92.677 

683  .  494 

17-5/8 

55.370 

243.977 

23-5/8 

74.220 

438.363 

29-5/8 

93.069 

689.298 

17-3/4 

55.763 

247.450 

23-3/4 

74.613 

443.014 

29-3/4 

93  .  462 

695.128 

17-7/8 

56.156 

250.947 

23-7/8 

75.005 

447.699 

29-7/8 

93.855 

700.981 

18 

56.548 

254.469 

24 

75.398 

452.390 

30 

94.248 

706.860 

18-1/8 

56.941 

258.016 

24-1/8 

75.791 

457.115 

30-1/8 

94.640 

712.762 

18-1/4 

57.334 

261.586 

24-1/4 

76.183 

461.864 

30-1/4 

95.033 

718.690 

18-3/8 

57.726 

265.182 

24-3/8 

76.576 

466.638 

30-3/8 

95.426 

724.641 

18-1/2 

58.119 

268.803 

24-1/2 

76.969 

471.436 

30-1/2 

95.818 

730.618 

18-5/8 

58.512 

272.447 

24-5/8 

77.361 

476.259 

30-5/8 

96.211 

736.619 

18-3/4 

58.905 

276.117 

24-3/4 

77.754 

481.106 

30-3/4 

96.604 

742.644 

18-7/8 

59  .  297 

279.811 

24-7/8 

78.147 

485.978 

30-7/8 

96.996 

748.694 

19 

59  .  690 

283  .  529 

25 

78.540 

490.875 

31 

97.389 

754  .  769 

19-1/8 

60.083 

287.272 

25-1/8 

78.932 

495.796 

31-1/8 

97.782 

760  .  868 

19-1/4 

60.475 

291.039 

25-1/4 

79  .  325 

500.741 

31-1/4 

98.175 

766.992 

19-3/8 

60.868 

294.831 

25-3/8 

79.718 

505.711 

31-3/8 

98.567 

773  .  140 

19-1/2 

61.261 

298.648 

25-1/2 

80.110 

510.706 

31-1/2 

98.968 

779.313 

19-5/8 

61.653 

302.489 

25-5/8 

80  .  503 

515.725 

31-5/8 

99  .  353 

785.510 

19-3/4 

62.046 

306.355 

25-3/4 

80.896 

520.769 

31-3/4 

99.745 

791.732 

19-7/8 

62.439 

310.245 

25-7/8 

81.288 

525.837 

31-7/8 

100.138 

797.978 

20 

62  .  832 

314.160 

26 

81.681 

530.930 

32 

100.531 

804.249 

20-1/8 

63  .  224 

318.099 

26-1/8 

82.074 

536.047 

32-1/8 

100.924 

810.545 

20-1/4 

63.617 

322.063 

26-1/4 

82.467 

541.189 

32-1/4 

101.316 

816.865 

20-3/8 

64.010 

326.051 

26-3/8 

82.859 

546.356 

32-378 

101.709 

823  .  209 

20-1/2 

64  .  402 

330  .  064 

26-1/2 

83.252 

551  .  547 

32-1/2 

102.102 

829.578 

20-5/8 

64  .  795 

334.101 

26-5/8 

83.645 

556.762 

32-5/8 

102.494 

835.972 

20-3/4 

65.188  338.163 

26-3/4 

84.037 

562.002 

32-3/4 

102.887 

842.390 

20-7/8 

65.580  1342.250! 

26-7/8 

84.430 

567.267: 

32-7/8 

103.280 

848.833 

270 


ELEMENTS  OF  HYDRAULICS 


CIRCUMFERENCES  AND  AREAS  OF  CIRCLES — (Continued} 


Diame- 
ter 
inches 

Circum- 
ference 
inches 

Area 
square 
inches 

Diame- 
ter 
inches 

Circum- 
ference 
inches 

Area 
square 
inches 

Diame- 
ter 
inches 

Circum- 
ference 
inches 

Area 
square 
inches 

33 

103.672 

855.30 

39 

122.522 

1194.59 

45 

141.372 

1590.43 

33-1/8 

104.055 

861  .  79 

39-1/8 

122.915 

1202.26 

45-1/8 

141.764 

1599.28 

33-1/4 

104  .  458 

868.30 

39-1/4 

123.307 

1209.95 

45-1/4 

142.157 

1608.15 

33-3/8 

104.850 

874.84 

39-3/8 

123.700 

1217.67 

45-3/8 

142.550 

1617.04 

33-1/2 

105.243 

881.41 

39-1/2 

124  .  093 

1225.42 

45-1/2 

142.942 

1625.97 

33-5/8 

105.636 

888.00 

39-5/8 

.124.485 

1233.18 

45-5/8 

143.335 

1634.92 

33-3/4 

106.029 

894.61 

39-3/4 

124.878 

1240.98 

45-3/4 

143.728 

1643.89 

33-7/8 

106.421 

901.25 

39-7/8 

125.271 

1248.79 

45-7/8 

144.120 

1652.88 

34 

106.814 

907.92 

40 

125.664 

1256.64 

46 

144.513 

1661.90 

34-1/8 

107.207 

914.61 

40-1/8 

126.056 

1264.50 

46-1/8 

144.906 

1670.95 

34-1/4 

107.599 

921.32 

40-1/4 

126.449 

1272.39 

46-1/4 

145.299 

1680.01 

34-3/8 

107.992 

928.06 

40-3/8 

126.842 

1280.31 

46-3/8 

145.691 

1689.10 

34-1/2 

108.385 

934.82 

40-1/2 

127.234 

1288.25 

46-1/2 

146.084 

1698.23 

34-5/8 

108.777 

941  .  60 

40-5/8 

127.627 

1296.21 

46-5/8 

146.477 

1707.37 

34-3/4 

109.170 

948.41 

40-3/4 

128.020 

1304  .  20 

46-3/4 

146.869 

1716.54 

34-7/8 

109.563 

955.25 

40-7/8 

128.412 

1312.21 

46-7/8 

147.262 

1725.73 

35 

109.956 

962.11 

41 

128.805 

1320.25 

47 

147.655 

1734.94 

35-1/8 

110.348 

968.99 

41-1/8 

129.198 

1328.32 

47-1/8 

148.047 

1744.18 

35-1/4 

110.741 

975.90 

41-1/4 

129.591 

1336.40 

47-1/4 

148.440 

1753.45 

35-3/8 

111.134 

982.84 

41-3/8 

129.983 

1344.51 

47-3/8 

148.833 

1762.73 

35-1/2 

111.526 

989  .  80 

41-1/2 

130.376 

1352.65 

47-1/2 

149  .  226 

1772.05 

35-5/8 

111.919 

996.78 

41-5/8 

130.769 

1360.81 

47-5/8 

149.618 

1781.39 

35-3/4 

112.312 

1003.78 

41-3/4 

131.161 

1369.00 

47-3/4 

150.011 

1790.76 

35-7/8 

112.704 

1010.82 

41-7/8 

131.554 

1377.21 

47-7/8 

150.404 

1800.14 

36 

113.097 

1017.88 

42 

131.947 

1385.44 

48 

150.796 

1809.56 

36-1/8 

113.490 

1024.95 

42-1/8 

132.339 

1393.70 

48-1/8 

151.189 

1818.99 

36-1/4 

113.883 

1032.06 

42-1/4 

132.732 

1401.98 

48-1/4 

151.582 

1828.46 

36-3/8 

114.275 

1039.19 

42-3/8 

133.125 

1410.29 

48-3/8 

151.974 

1837.93 

36-1/2 

114.668 

1046.35 

42-1/2 

133.518 

1418.62 

48-1/2 

152.367 

1847.45 

36-5/8 

115.061 

1053.52 

42-5/8 

133.910 

1426.98 

48-5/8 

152.760 

1856.99 

36-3/4 

115.453 

1060.73 

42-3/4 

134.303 

1435.36 

48-3/4 

153.153 

1866.55 

36-7/8 

115.846 

1067.95 

42-7/8 

134.696 

1443.77 

48-7/8 

153.545 

1876.13 

37 

116.239 

1075.21 

43 

135.088 

1452.20 

49 

153.938 

1885.74 

37-1/8 

116.631 

1082.48 

43-1/8 

135.481 

1460.65 

49-1/8 

154.331 

1895.37 

37-  1/4 

117.024 

1089  .  79 

43-1/4 

135.874 

1469.13 

49-1/4 

154.723 

1905.03 

37-3/8 

117.417 

1097.11 

43-3/8 

136.266 

1477.63 

49-3/8 

155.116 

1914.70 

37-1/2 

117.810 

1104.46 

43-1/2 

136.659 

1486.17 

49-1/2 

155.509 

1924.42 

37-5/8 

118.202 

1111.84 

43-5/8 

137.052 

1494.72 

49-5/8 

155.901 

1934.15 

37-3/4 

118.595 

1119.24 

43-3/4 

137.445 

1503.30 

49-3/4 

156.294 

1943.91 

37-7/8 

118.988 

1126.66 

43-7/8 

137.837 

1511.90 

49-7/8 

156.687 

1953.69 

38 

119.380 

1134.11 

44 

138.230 

1520.53 

50 

157.080 

1963.50 

38-1/8 

119.773 

1141.59 

44-1/8 

138.623 

1529.18 

50-1/4 

157.865 

1983.18 

38-1/4 

120.166 

1149.08 

44-1/4 

139.015 

1537.86 

50-1/2 

158.650 

2002.96 

38-3/8 

120.558 

1156.61 

44-3/8 

139.408 

1546.55 

50-3/4 

159.436 

2022  .  84 

38-1/2 

120.951 

1164.15 

44-1/2 

139.801 

1555.28 

51 

160.221 

2042  .  82 

38-5/8 

121.344 

1171.73 

44-5/8 

140.193 

1564.03 

51-1/4 

161.007 

2062  .  90 

38-3/4 

121.737 

1179.32  1|  44-3/4 

140.586 

1572.81 

51-1/2 

161.792 

2083  .  07 

38-7/8 

122.129 

1186.  94!  |  44-7/8 

140.979 

1581.61 

51-3/4 

162.577 

2103.35 

TABLES 


271 


CIRCUMFERENCES  AND  AREAS  OF  CIRCLES — (Continued) 


Diame- 
ter 
inches 

Circum- 
ference 
inches 

Area 
square 
inches 

Diame- 
ter 
inches 

Circum- 
ference 
inches 

Area 
square 
inches 

Diame- 
ter 
inches 

Circum- 
ference 
inches 

Area 
square 
inches 

52 

163.363 

2123.72 

63 

197.920 

3117.25 

74 

232.478 

4300  .  84 

52-1/4 

164.148 

2144.19 

63-1/4 

198.706 

3142.04 

74-1/4 

233.263 

4329.95 

52-1/2 

164.934 

2164.75 

63-1/2 

199.491 

3166.92 

74-1/2 

234.049 

4359.16 

52-3/4 

165.719 

2185.42 

63-3/4 

200.277 

3191.91 

74-3/4 

234  .  834 

4388  .  47 

53 

166.504 

2206.18 

64 

201.062 

3216.99 

75 

235.620 

4417.86 

53-1/4 

167.490 

2227.05 

64-1/4 

201  .  847 

3242.17 

75-1/4 

236.405 

4447.37 

53-1/2 

168.075 

2248.01 

64-1/2 

202.633 

3267.46 

75-1/2 

237.190 

4476.97 

53-3/4 

168.861 

2269.06 

64-3/4 

203.418 

3292.83 

75-3/4 

237.976 

4506.67 

54- 

169.646 

2290  .  22 

65 

204.204 

3318.31 

76 

238.761 

4536.46 

54-1/4 

170.431 

2311.48 

65-1/4 

204.989 

3343.88 

76-1/4 

239.547 

4566.36 

54-1/2 

171.217 

2332.83 

65-1/2 

205.774 

3369.56 

76-1/2 

240.332 

4596.35 

54-3/4 

172.002 

2354.28 

64-3/4 

206.560 

3395.33 

76-3/4 

241.117 

4626.44 

55 

172.788 

2375.83 

66 

207.345 

3421  .  19 

77 

241.903 

4656.63 

55-1/4 

173.573 

2397.48 

66-1/4 

208.131 

3447.16 

77-1/4 

242.688 

4686.92 

55-1/2 

174.358 

2419.22 

66-1/2 

208.916 

3473.33 

77-1/2 

243.474 

4717.30 

55-3/4 

175.144 

2441.07 

66-3/4 

209.701 

3499.39 

77-3/4 

244.259 

4747.79 

56 

175.929 

2463.01 

67 

210.487 

3525  .  66 

78 

245.044 

4778.36 

56-1/4 

176.715 

2485.05 

67-1/4 

211.272 

3552.01 

78-1/4 

245  .  830 

4809.05 

56-1/2 

177.500 

2507.19 

67-1/2 

212.058 

3578.47 

78-1/2 

246.615 

4839.83 

56-3/4 

178.285 

2529.42 

67-3/4 

212.843 

3605.03 

78-3/4 

247.401 

4870.70 

57 

179.071 

2551  .  76 

68 

213.628 

3631.68 

79 

248.186 

4901  .  68 

57-1/4 

179.856 

2574.19 

68-1/4 

214.414 

3658.44 

79-1/4 

248.971 

4932.75 

57-1/2 

180.642 

2596.72 

68-1/2 

215.199 

3685.29 

79-1/2 

249  .  757 

4963.92 

57-3/4 

181.427 

2619.35 

68-3/4 

215.985 

3712.24 

79-3/4 

250.542 

4995.19 

58 

182.212 

2642.08 

69 

216.770 

3739.28 

80 

251  .  328 

5026.55 

58-1/4 

182.998 

2664.91 

69-1/4 

217.555 

3766.43 

80-1/2 

252.898 

5089.58 

58-1/2 

183.783 

2687.83 

69-1/2 

218.341 

3793.67 

58-3/4 

184.569 

2710.85 

69-3/4 

219.126 

3821.02 

81 

254.469 

5153.00 

81-1/2 

256  .  040 

5216.82 

59 

185.354 

2733.97 

70 

219.912 

3848.45 

82 

257.611 

5281.02 

59-1/4 

59-1/2 

186.139 
186.925 

2757.19 
2780.51 

70-1/4 
70-1/2 

220.697 
221.482 

3875.99 
3903.63 

82-1/2 

259.182 

5345.62 

59-3/4 

187.710 

2803.92 

70-3/4 

222.268 

3931.36 

83 

260.752 

5410.61 

83-1/2 

262  .  323 

5476.00 

60 

188.496 

2827.43 

71 

223.053 

3959.19 

60-1/4 

189.281 

2851.05 

71-1/4 

223.839 

3987.13 

84 

263.894 

5541.77 

60-1/2 

190.066 

2874  .  76 

71-1/2 

224  .  624 

4015.16 

84-1/2 

265.465 

5607  .  95 

60-3/4 

190.852 

2898.56 

71-3/4 

225  .  409 

4043  .  28 

85 

267.035 

5674.51 

61 

191.637 

2922.47 

72 

226.195 

4071  .  50 

85-1/2 

268.606 

5741.47 

61-1/4 

192.423 

2946.47 

72-1/4 

226.980 

4099  .  83 

86 

270.  177 

5808  .  80 

61-1/2 
61-3/4 

193  .  208 
193.993 

2970.57 
2994.77 

72-1/2 
72-3/4 

227.766 
228.551 

4128.25 
4156.77 

86-1/2 

271.748 

5876.55 

87 

273.319 

5944  .  68 

62 

194  .  779 

3019.07 

73 

229.336 

4185.39 

87-1/2 

274.890 

6013.21 

62-1/4 

195.564 

3043.47 

73-1/4 

230.122 

4214.11 

62-1/2 

196.350 

3067  .  96 

73-1/2 

230.907 

4242.92 

88 

276.460 

6082.12 

62-3/4 

197.135 

3092.56 

73-3/4 

231.693 

4271.83  II  88-1/2 

278.031 

6151.44 

272 


ELEMENTS  OF  HYDRAULICS 


CIRCUMFERENCES  AND  AREAS  OF  CIRCLES — (Continued} 


Diame- 
ter 
inches 

Circum- 
ference 
inches 

Area 
square 
inches 

Diame- 
ter 
inches 

Circum- 
ference 
inches 

Area 
square 
inches 

Diame- 
ter 
inches 

Circum- 
ference 
inches 

Area 
square 
inches 

89 
89-1/2 

279.602 
281.173 

6221.14 
6291.25 

100 
100-1/2 

314.159 
315.730 

7853.98 
7938.72 

111 
111-1/2 

348.717 
350.288 

9766.89 
9674  .  28 

90 
90-1/2 

282  .  744 
284.314 

6361  .  73 
6432.62 

101 

101-1/2 

317.301 

318.872 

8011.85 
8091  .  36 

112 

112-1/2 

351.858 
353.430 

9852.03 
9940  .  20 

91 
91-1/2 

285.885 
287.456 

6503.88 
6573.56 

102 
102-1/2 

320.442 
322.014 

8171.28 
8251.60 

113 
113-1/2 

355  .  000 
356.570 

10028.75 
10117.68 

92 
92-1/2 

289  .  027 
290  .  598 

6647.61 
6720.07 

103 
103-1/2 

323  .  584 
325.154 

8332.29 
8413.40 

114 
114-1/2 

358.142 
359.712 

10207.03 
10296.76 

93 
93-1/2 

292.168 
293  .  739 

6792.91 
6866.16 

104 
104-1/2 

326.726 
328.296 

8494.87 
8576.76 

115 
115-1/2 

361  .  283 
362.854 

10386  .  89 
10477  .  40 

94 
94-1/2 

295.310 

296.881 

6939  .  78 
7013.81 

105 
105-1/2 

329.867 
331.438 

8659.01 
8741.68 

116 
116-1/2 

364.425 
365.996 

10568.32 
10659.64 

95 
95-1/2 

298.452 
300.022 

7088.22 
7163.04 

106 
106-1/2 

333.009 
334  .  580 

8824.73 
8908.20 

117 
117-1/2 

367.566 
369  .  138 

10751.32 
10843  .  40 

96 
96-1/2 

301  .  593 
302.164 

7238.  23 
7313.84 

107 
107-1/2 

336.150 
337.722 

8992.02 
9076.24 

118 
118-1/2 

370.708 
372.278 

10935.88 
11028.76 

97 
97-1/2 

304.734 
306.306 

7389.81 
7474.20 

108 
108-1/2 

339  .  292 
340  .  862 

9160.88 
9245.92 

119 
119-1/2 

373.849 
375.420 

11122.02 

11215.68 

98 

98-1/2 

307.876 
309.446 

7542.96 
7620.  12 

109 
109-1/2 

342.434 
344  .  004 

9331.32 
9417.12 

120 

376.991 

11309.73 

99 

311.018 

7697.69 

110 

345.575 

9503.32 

99-1/2 

312.588 

7775.64 

110-1/2 

347.146 

9589.92 

TABLES 


273 


TABLE  9.1 — EFFLUX  COEFFICIENTS  FOR  CIRCULAR  ORIFICE 

Values  of  efflux  coefficient  K  in  Eq.  (25),  Art.  8,  Q  =  2/3Kb\/2g(H3/2  -  fc3/2),  for  circular, 

vertical  orifices,  with  sharp  edges,  full  contraction  and  free  discharge  in  air. 

For  heads  over  100  ft.,  use  K  =  0.592. 


Head 
on  cen- 
ter of 
orifice 
in  feet 

Diameter  of  orifice  in  feet 

0.02 

0.03 

0.04 

0.05 

0.07 

0.10 

0.12 

0.15 

0.20 

0.40 

0.60 

0.80 

1.0 

0.3 

0.4 
0.5 
0.6 
0.7 

0.637 

0  6280.621 

0  613 

0  608 

0.655 
0.651 

0.643 
0.640 
0.637 

0.637 
0.633 
0.630 
0.628 

0.631 
0.627 
0.624 
0.622 

0.624 
0.621 
0.618 
0.616 

0.618 
0.615 
0.613 
0.611 

0.612 
0.610 
0.609 
0.607 

0.606 
0.605 
0.605 
0.604 

0.600 
0.601 
0.601 

0.596 
0.596 
0.597 

0.592 
0.593 
0.594 

0.590 
0.591 

0.590 

0.8 
0.9 
1.0 
1.2 
1.4 

0.648 
0.646 
0.644 
0.641 
0.638 

0.634 
0.632 
0.631 
0.628 
0.625 

0.626 
0.624 
0.623 
0.620 
0.618 

0.620 
0.618 
0.617 
0.615 
0.613 

0.615 
0.613 
0.612 
0.610 
0.609 

0.610 
0.609 
0.608 
0.606 
0.605 

0.606 
0.605 
0.605 
0.604 
0.603 

0.603 
0.603 
0.603 
0.602 
0.601 

0.601 
0.601 
0.600 
0.600 
0.600 

0.597 
0.598 
0.598 
0.598 
0.599 

0.594 
0.595 
0.595 
0.596 
0.596 

0.592 
0.593 
0.593 
0.594 
0.594 

0.591 
0.591 
0.591 
0.592 
0.593 

1.6 
1.8 
2.0 
2.5 
3.0 

0.636 
0.634 
0.632 
0.629 
0.627 

0.624 
0.622 
0.621 
0.619 
0.617 

0.617 
0.615 
0.614 
0.612 
0.611 

0.612 
0.611 
0.610 
0.608 
0.606 

0.608 
0.607 
0.607 
0.605 
0.604 

0.605 
0.604 
0.604 
0.603 
0.603 

0.602 
0.602 
0.601 
0.601 
0.601 

0.601 
0.601 
0.600 
0.600 
0.600 

0.600 
0.599 
0.599 
0.599 
0.599 

0.599 
0.599 
0.599 
0.599 
0.599 

0.5970.595 
0.5970.595 
0.5970.596 
0.5980.597 
0.5980.597 

0.594 
0.595 
0.595 
0.596 
0.597 

3.5 

4.0 
5.0 
6.0 
7.0 

0.625 
0.623 
0.621 
0.618 
0.616 

0.616 
0.614 
0.613 
0.611 
0.609 

0.610 
0.609 
0.608 
0.607 
0.606 

0.606 
0.605 
0.605 
0.604 
0.603 

0.604 
0.603 
0.603 
0.602 
0.601 

0.602 
0.602 
0.601 
0.600 
0.600 

0.601 
0.600 
0.599 
0.599 
0.599 

0.600 
0.599 
0.599 
0.599 
0.599 

0.599 
0.599 
0.598 
0.598 
0.598 

0.599 
0.598 
0.598 
0.598 
0.958 

0.598 
0.597 
0.597 
0.597 
0.597 

0.597 
0.597 
0.596 
0.596 
0.596 

0.596 
0.596 
0.596 
0.596 
0.596 

8.0 
9.0 
10.0 
20.0 
50.0 
100.0 

0.614 
0.613 
0.611 
0.601 
0.596 
0.593 

0.608 
0.607 
0.606 
0.600 
0.596 
0.593 

0.605 
0.604 
0.603 
0.599 
0.595 
0.592 

0.603 
0.602 
0.601 
0.598 
0.595 
0.592 

0.601 
0.600 
0.599 
0.597 
0.594 
0.592 

0.600 
0.599 
0.598 
0.596 
0.594 
0.592 

0.599 
0.599 
0.598 
0.596 
0.594 
0.592 

0.598 
0.598 
0.597 
0.596 
0.594 
0.592 

0.5980.597 
0.5970.597 
0.5970.597 
0.5960.596 
0.5940.594 
0.5920.592 

0.596 
0.596 
0.596 
0.596 
0.594 
0.592 

0.596 
0.596 
0.596 
0.595 
0.593 
0.592 

0.596 
0.595 
0.595 
0.594 
0.593 
0.592 

1  From  Hamilton  Smith's  Hydraulics. 


18 


274 


'    ELEMENTS  OF  HYDRAULICS 


TABLE  10. l — EFFLUX  COEFFICIENTS  FOR  SQUARE  ORIFICE 

Values  of  efflux  coefficient  K  in  Eq.  (25),  Art.  8,  Q  =  2/3  Kb\/2-j(Hz'2-  h3/2),  for  square, 

vertical  orifices,  with  sharp  edges,  full  contraction,  and  free  discharge  in  air. 

For  heads  over  100  ft.,  use  K  =  0.598 


Head 
on  cen- 
ter of 
orifice 
in  feet 

Side  of  square  in  feet 

0.02 

0.03 

0.04 

0.50 

0.07 

0.10 

0.12 

0.15 

0.20 

0.40 

0.60 

0.80 

1.0 

0.3 
0.4 
0.5 
0.6 
0.7 

0.660 
0.656 

0  .  648 
0.645 
0.642 

0.643 
0.639 
0.636 
0.633 

0.642 
0.637 
0.633 
0.630 
0.628 

0.632 
0.628 
0.625 
0.623 
0.621 

0.624 
0.621 
0.619 
0.617 
0.616 

0.617 
0.616 
0.614 
0.613 
0.612 

0.612 
0.611 
0.610 
0.610 
0.609 

0.605 
0.605 
0.605 

0.601  0.597 
0.601  0.598 
0.6020.599 

0.596 
0.598 

0.596 

0.8 
0.9 
.0 
.2 
.4 

0.652 
0.650 
0.648 
0.644 
0.642 

0.639 
0.637 
0.636 
0.623 
0.630 

0.631 
0.629 
0.628 
0.625 
0.623 

0.625 
0.623 
0.622 
0.620 
0.618 

0.620 
0.619 
0.618 
0.616 
0.614 

0.615 
0.614 
0.613 
0.611 
0.610 

0.611 

0.610 
0.610 
0.609 
0.608 

0.608 
0.608 
0.608 
0.607 
0.606 

0.605 
0.605 
0.605 
0.605 
0.605 

0.6020.6000.598 
0.603!0.601  0.599 
0.6030.601j0.600 
0.6040.6020.601 
0.6040.6020.601 

0.597 
0.598 
0.599 
0.600 
0.601 

.6 
.8 
2.0 
2.5 
3.0 

0.640 
0.638 
0.637 
0.634 
0.632 

0.628 
0.627 
0.626 
0.624 
0.622 

0.621 
0.620 
0.619 
0.617 
0.616 

0.617 
0.616 
0.615 
0.613 
0.612 

0.613 
0.612 
0.612 
0.610 
0.609 

0.609 
0.609 
0.608 
0.607 
0.607 

0.607 
0.607 
0.606 
0.606 
0.606 

0.606 
0.606 
0.605 
0.605 
0.605 

0.606 
0.606 
0.606 
0.606 
0.606 

0.605 
0.605 
0.605 
0.605 
0.605 

0.6050.6030.602 
0.6050.603!0.602 
0.6050.6040.602 
0.605^.6040.603 
0.605'Q.  604  0.603 

0.601 
0.602 
0.602 
0.602 
0.603 

3.5 
4.0 
5.0 
6.0 
7.0 

0.630 
0.628 
0.626 
0.623 
0.621 

0.621 
0.619 
0.617 
0.616 
0.615 

0.615 
0.614 
0.613 
0.612 
0.611 

0.611 
0.610 
0.610 
0.609 
0.608 

0.609 
0.608 
0.607 
0.607 
0.607 

0.607 
0.606 
0.606 
0.605 
0.605 

0.605 
0.605 
0.605 
0.605 
0.604 

0.605 
0.605 
0.604 
0.604 
0.604 

0.605 
0.605 
0.604 
0.604 
0.604 

0.604 
0.603 
0.603 
0.603 
0.603 

0.6030.602 
0.6030.602 
0.6020.602 
0.6020.602 
0.6020.602 

8.0 
9.0 
10.0 
20.0 
50.0 
100.0 

0.619 
0.618 
0.616 
0.606 
0.602 
0.599 

0.613 
0.612 
0.611 
0.605 
0.601 
0.598 

0.610 
0.609 
0.608 
0.604 
0.601 
0.598 

0.608 
0.607 
0.606 
0.603 
0.601 
0.598 

0.606 
0.606 
0.605 
0.602 
0.601 
0.598 

0.605 
0.604 
0.604 
0.602 
0.600 
0.598 

0.604 
0.604 
0.604 
0.602 
0.600 
0.598 

0.604 
0.604 
0.603 
0.602 
0.600 
0.598 

0.604 
0.603 
0.603 
0.602 
0.600 
0.598 

0.603 
0.603 
0.603 
0.601 
0.600 
0.598 

0.603 
0.602 
0.602 
0.601 
0.599 
0.598 

0.602 
0.602 
0.602 
0.601 
0.599 
0.598 

0.602 
0.601 
0.601 
0.600 
0.599 
0.598 

1  From  Hamilton  Smith's  Hydraulics. 


TABLES 


275 


TABLE  11. — FIRE  STREAMS 

From  Tables  Published  by  John  R.  Freeman 


3/4-in.  Smooth  Nozzle 

Pressure  in  pounds  per  sq.  in.  re- 

Pressure at 
nozzle  in 
pounds  per 

Discharge  in 
gallons  per 
min. 

Height  of 
effective 
fire  stream 

Horizontal 
distance 
of  stream 

quired  at  hydrant  or  pump  to  main- 
tain pressure  at  nozzle  through  vari- 
ous lengths  of  2-1/2-in.  smooth, 
rubber-lined  hose. 

sq.  in. 

50 

100 

200 

300400 

500 

600 

800 

1000 

ft. 

ft. 

ft. 

ft. 

ft. 

ft. 

ft. 

ft. 

ft. 

35 

97 

55 

41 

37 

38 

40 

42 

44 

46 

48 

53 

57 

40 

104 

60 

44 

42 

43 

46 

48 

50 

53 

55 

60 

65 

45 

110 

64 

47 

47 

48 

51 

54 

57 

59 

62 

68 

73 

50 

116 

67 

50 

52 

54 

57 

60 

63 

66 

69 

75 

81 

55 

122 

70 

52 

58 

59 

63 

66 

69 

73 

76 

83 

89 

60 

127 

72 

54 

63 

65 

68 

72 

76 

79 

83 

90 

97 

65 

132 

74 

56 

68 

70 

74 

78 

82 

86 

90 

98 

106 

70 

137 

76 

58 

73 

75 

80 

84 

88 

92 

97 

105 

114 

75 

142 

78 

60 

79 

81 

85 

90 

94 

99 

104 

113 

122 

80 

147 

79 

62 

84 

86 

91 

96 

101 

106 

1.11 

120 

130 

85 

151 

80 

64 

89 

92 

97 

102 

107 

112 

117 

128 

138 

90 

156 

81 

65 

94 

97 

102 

108 

113 

119 

124 

135 

146 

95 

160 

82 

66 

99 

102 

108 

114 

120 

125 

131 

143 

154 

100 

164 

83 

68 

105 

108 

114 

120 

126 

132 

138 

150 

163 

7/8-in.  Smooth  Nozzle 

35 

133 

56 

46 

38 

40 

44 

48 

52 

56 

60 

68 

76 

40 

142 

62 

49 

43 

46 

50 

55 

59 

64 

68 

78 

87 

45 

150 

67 

52 

49 

51 

57 

62 

67 

72 

77 

87 

97 

50 

159 

71 

55 

54 

57 

63 

69 

74 

80 

86 

97 

108 

55 

166 

74 

58 

60 

63 

69 

75 

82 

88 

94 

107 

119 

60 

174 

77 

61 

65 

69 

75 

82 

89 

96 

103 

116 

130 

65 

181 

79 

64 

71 

74 

82 

89 

96 

104 

111 

126 

141 

70 

188 

81 

66 

76 

80 

88 

96 

104 

112 

120 

136 

152 

75 

194 

83 

68 

82 

86 

94 

103 

111 

120 

128 

145 

162 

80 

201 

85 

70 

87 

91 

101 

110 

119 

128 

137 

155 

173 

85 

207 

87 

72 

92 

97 

107 

116 

126 

136 

145 

165 

184 

90 

213 

88 

74 

98 

103 

113 

123 

134 

144 

154 

174 

195 

95 

219 

89 

75 

103 

109 

119 

130 

141 

152 

163 

184 

206 

100 

224 

90 

76 

109 

114 

126  137 

148 

160 

171 

194 

216 

1-in.  Smooth  Nozzle 

35 

174 

58 

51 

40 

44 

51 

57 

64 

71 

78 

92 

105 

40 

186 

64 

55 

46 

50 

58 

66 

73 

81 

89 

105 

120 

45 

198 

69 

58 

52 

56 

65 

74 

83 

91 

100 

118 

135 

'  50 

208 

73 

61 

57 

62 

72 

82 

92 

102 

111 

131 

151 

55 

218 

76 

64 

63 

69 

79 

90 

101 

112 

122 

144 

166 

60 

228 

79 

67 

67 

75 

87 

98 

110 

122 

134 

157 

181 

65 

237 

82 

70 

75 

81 

94 

107 

119 

132 

145 

170 

196 

70 

246 

85 

72 

80 

87 

101 

115 

128 

142 

156 

183 

211 

75 

255 

87 

74 

86 

94 

110 

123 

138 

152 

167 

196 

226 

80 

263 

89 

76 

92 

100 

115 

131 

147 

162 

178 

209 

241 

85 

274 

91 

78 

98 

106 

123 

139 

156 

173 

189 

222 

90 

279 

92 

80 

103 

112 

130 

147 

165 

183 

200 

236 

95 

287 

94 

82 

109 

118 

137 

156 

174 

193 

211 

249 

100 

295 

96 

83 

115 

125 

144 

164 

183 

203 

223 

276 


ELEMENTS  OF  HYDRAULICS 


FIRE  STREAMS — (Continued} 


1-1/8-inch  Smooth  Nozzle 

% 

Pressure  in  pounds  per  sq.  in.  required 

Pressure  at 
nozzle  in 
pound  per 
sq.  in. 

Discharge  in 
gallons  per 
min. 

Height  of 
effective 
fire  stream 

Horizontal 
distance 
of  stream 

at  hydrant  or  pump  to  maintain  pres- 
sure at  nozzle  through  various  lengths 
of  2-1/2-in.  smooth,  rubber-lined  hose 

50 

100 

200 

300 

400 

500 

600 

800  jll  000 

ft. 

ft. 

ft. 

ft. 

ft. 

ft. 

ft. 

ft. 

ft. 

35 

222 

59 

54 

43 

49 

60 

71 

82 

94 

105 

127 

149 

40 

238 

65 

59 

50 

56 

69 

81 

94 

107 

120 

145 

171 

45 

252 

70 

63 

56 

63 

77 

92 

106 

120 

135 

163 

192 

50 

266 

75 

66 

62 

70 

86 

102 

118 

134 

150 

181 

213 

55 

279 

80 

69 

68 

77 

95 

112 

130 

147 

165200 

235 

60 

291 

83 

72 

74 

84 

103 

122 

141 

160 

180 

218 

256 

65 

303 

86 

75 

81 

91 

112 

132 

153 

174 

195 

236 

70 

314 

88 

77 

87 

98 

120 

143 

165 

187 

209 

254 

75 

325 

90 

79 

93 

105 

12,9 

153 

177 

201 

224 

80 

336 

92 

81 

99 

112 

138 

163 

188 

214 

239 

85 

346 

94 

83 

106 

119 

146 

173 

200 

227 

254 

90 

356 

96 

85 

112 

126 

15& 

183 

212 

241 

95 

366 

98 

87 

118 

133 

163 

194 

224 

254 

... 

100 

376 

99 

89 

124 

140 

172 

204 

236 

1-1/4-inch  Smooth  Nozzle 

35 

277 

60 

59 

48 

57 

74 

91 

109 

126 

142 

178 

212 

40 

296 

67 

63 

55 

65 

84 

104 

124 

144 

164 

203 

243 

45 

314 

72 

67 

62 

73 

95 

117 

140 

162 

184 

229 

50 

331 

77 

70 

68 

81 

106 

130 

155 

180 

204 

254 

55 

347 

81 

73 

75 

89 

116 

143 

170 

198 

225 

60 

363 

85 

76 

82 

97 

127 

156 

186 

216 

245 

65 

377 

88 

79 

89 

105 

137 

169 

201 

234 

70 

392 

91 

81 

96 

113 

148 

182 

217 

252 

75 

405 

93 

83 

103 

121 

158 

195 

232 

80 

419 

95 

85 

110 

129 

169 

208 

248 

85 

432 

97 

88 

116 

137 

179 

221 

90 

444 

99 

90 

123 

145 

190 

234 

95 

456 

100 

92 

130 

154 

210 

247 

100 

468 

101 

93 

137 

162 

211 

261 

1-3/8-inch  Smooth  Nozzle 

35 

340 

62 

62 

54 

67 

94 

120 

146 

172 

198 

250 

40 

363 

69 

66 

62 

77 

107 

137 

166 

196 

226 

45 

385 

74 

70 

70 

87 

120 

154 

187 

221 

454 

50 

406 

79 

73 

78 

96 

134 

171 

208 

245 

55 

426 

83 

76 

86 

106 

147 

188 

229 

270 

60 

445 

87 

79 

93 

116 

160 

205 

250 

65 

463 

90 

82 

101 

125 

174 

222 

70 

480 

92 

84 

109 

135 

187 

239 

75 

497 

95 

86 

117 

145 

201 

256 

on 

514 

97 

88 

124 

154 

214 

oU 

85 

529 

99 

90 

132 

164 

227 

90 

545 

100 

92 

140 

173 

240 

95 

560 

101 

94 

148 

183 

254 

100 

574 

103 

96 

156 

193 

TABLES 


277 


TABLE  12.  l  —  COEFFICIENTS  OF  PIPE  FRICTION 

Values  of  the  friction  coefficient,  /,  in  the  formula 


Computed  from  the  exponential  formulas  of  Thrupp,  Tutton  and  Unwin 


Diameter 

' 

/elocity  of 

flow  in  feet 

per  second 

in  inches 

2 

4 

6 

8 

10 

Lead  pipe 

1 
2 
3 
4 

0.032 
0.030 
0.029 
0.028 

0.026 
0.025 
0.024 
0.023 

0.024 
0.023 
0.022 
0.021 

0.022 
0.021 
0.020 
0.020 

0.021 
0.020 
0.019 
0.019 

6 

0.034 

0.033 

0.032 

0.032 

12 
18 

0.027 
0  024 

0.027 
0  024 

0.026 
0  023 

0.026 
0  023 

24 
36 

48 

0.022 
0.020 
0  018 

0.022 
0.019 
0  018 

0.021 
0.019 
0  017 

0.021 
0.019 
0  017 

Asphalted  pipe 

6 
9 
12 

18 
24 
36 

48 

0.026 
0.025 
0.024 
0.023 
0.022 
0.021 
0.020 

0.023 
0.022 
0.021 
0.020 
0.020 
0.019 
0.018 

0.022 
0.021 
0.020 
0.019 
0.018 
0.017 
0.017 

0.021 
0.020 
0.019 
0.018 
0.017 
0.017 
0.016 

0.020 
0.019 
0.019 
a.  018 
0.017 
0.016 
0.015 

Bare  wrought  iron 
pipe 

3 
6 
12 
24 
36 
48 
60 

0.024 
0.022 
0.019 
0.017 
0.016 
0.015 
0.015 

0.021 
0.019 
0.017 
0.015 
0.014 
0.013 
0.013 

0.019 
0.017 
0.015 
0.014 
0.013 
0.012 
0.012 

0.018 
0.016 
0.014 
0.013 
0.012 
0.011 
0.011 

0.017 
0.016 
0.014 
0.012 
0.011 
0.011 
0.010 

Riveted  wrought  iron 
or  steel  pipe 

12 
24 
36 

48 
60 

72 

0.025 
0.020 
0.017 
0.016 
0.015 
0.014 

0.022 
0.018 
0.016 
0.014 
0.013 
0.013 

0.021 
0.017 
0.015 
0.014 
0.013 
0.012 

0.020 
0.016 
0.014 
0.013 
0.012 
0.011 

0.019 
0.016 
0.014 
0.013 
0.012 
0.011 

3 

6 

0.028 
0  024 

0.026 
0  022 

0.025 
0  022 

0.025 
0  021 

9 

0.021 

0.020 

0.020 

0.019 

New  cast-iron  pipe 

12 

18 
24 
36 

0.020 
0.018 
0.017 
0  015 

0.019 
0.017 
0.016 
0  015 

0.018 
0.017 
0.016 
0  014 

0.018 
0.016 
0.015 
0  014 

••••••••• 

3 

6 

0.059 
0  050 

0.058 
0  050 

0.058 
0  050 

0.058 
0  049 

Old  cast  iron  pipe 

9 
12 
18 
24 

36 

0.046 
0.043 
0.039 
0.037 
0.033 

0.045 
0.042 
0.039 
0.036 
0.033 

0.045 
0.042 
0.038 
0.036 
0.033 

0.044 
0.042 
0.038 
0.036 
0.032 

••••••••• 

1  Compiled  from  data  in  Gibson's  Hydraulics 


278 


ELEMENTS  OF  HYDRAULICS 


TABLE  13. — FRICTION  HEAD  IN  PIPES 

Friction  head  in  feet  for  each  100  ft.  of  straight,  clean,  cast-iron  pipe.  For  old  pipes  the 
tabular  values  of  the  friction  head  should  be  doubled.  Computed  from  Williams  and 
Hazen's  formula,  »  =  Cr°-«ss°-54  0.001 -°'04:  »  =  velocity  in  feet  per  sec.,  s  =  slope; 
r  =  hydraulic  radius  in  feet,  C  =  100. 


c  'a 

tk 

II 

Inside  diameter  of  pipe 

1/2  in. 

3/4  in. 

1  in. 

1-1/2  in. 

2  in. 

Velocity  in  feet 
per  sec. 

Friction  head  in 
feet  per  100  it. 

Velocity  in  ft. 
per  sec. 

ll« 

Velocity  in  feet 
per  seo. 

|| 

.3  g 

en 

fel 

!.] 

§  Id 

1*8 

a  d 
.-a  a 

O    "^ 

'c  c  2 

fc  . 

>  * 

fe          If  ^ 

i 

2 
3 
4 
5 

1.05 
2.10 
3.16 
4.21 
5.26 

2.1 
7.4 
15.8 
27.0 
41.0 

i| 

1.20 
1.80 
2.41 
3.01 

1.9 
4.1 
7.0 
10.5 

1.12 
1.49 
1.86 

1.26 
2.14 
3.25 

0.63 
0.79 

0.26 
0.40 



i  0.61 
0.82 
1.02 
1.23 
1.53 

6 
8 
10 
12 
15 

6.31 
8.42 
10.52 

57.0 
98.0 
147.0 

3.61 
4.81 
6.02 

7.22 
9.02 

14.7 
25.0 
38.0 
53.0 
80.0 

2.23 
2.98 
3.72 
4.46 
5.57 

4.55 
7.8 
11.7 
16.4 
25.0 

0.94 
1.26 
1.57 
1.89 
2.36 

0.56 
0.95 
1.43 
2.01 
3.05 

0.20 
0.33 
0.50 
0.70 
1.07 

20 
25 
30 
35 
40 

12.03 

136.0 

7.44 
9.30 
11.15 
13.02 
14.88 

42.0 
64.0 
89.0 
119.0 
152.0 

3.15 
3.93 
4.72 
5.51 
6.30 

5.2 
7.8 
11.0 
14.7 
18.8 

2.04 
2.55 
3.06 
3.57 
4.08 

1.82 
2.73 
3.84 
5.1 
6.6 

50 
60 
70 
80 
90 

7.87 
9.44 
11.02 
12.59' 
14.17 

28.4 
39.6 
53.0 
68.0 
84.0 

5.11 
6.13 
7.15 
8.17 
9.19 

9.9 
13.9 
18.4 
23.7 
29.4 

100 
120 
140 
160 
180 

! 



15.74 
18.89 
22.04 

102.0 
143.0 
190.0 

10.21 
12.25 
14.30 
16.34 
18.38 

35.8 
50.0 
67.0 
86.0 
107.0 

|  

200 
250 

1                                  i 

20.42 
25.53 

129.0 
196.0 

1  

TAKLKS 


FRICTION  HEAD  IN  PIPES — (Continued) 


Discharge  in 
gallons  per  min. 

Insult'  ili:imt'tt>r  of  pipe 

2-1/2  in. 

3  in. 

4  in. 

5  in. 

6  in. 

•a  I 
gS 

1.1 

Friction  head  in 
feet  per  100  ft. 

Velocity  in  feet 
per  sec. 

Jl 

JJU 

!*§ 

•al 

11 

I» 

Jlfi 

i"§ 

Velocity  in 
feet  per  sec. 

Friction  head 
in  feet  per 
100ft. 

•si 

t* 

|l 

|i 

|J* 

f.S§ 

10 
15 
20 
25 
30 

0.65 
0.98 
1.31 
1.63 
1.96 

0.17 
0.37 
0.61 
'    0.92 
1.29 

0.45 
0.68 
0.91 
1.13 
1.36 

0.07 
0.15 
0.25 
0.38 
0.54 

0.51 
0.64 
0.77 

0.06 
0.09 
0.13 

0.49 

0.04 

35 
40 
50 
60 
70 

2.29 
2.61 
3.27 
3.92 
4.58 

1.72 
2.20 
3.32 
4.65 
6.2 

1.59 
1.82 
2.27 
2.72 
3.18 

0.71 
0.91 
1.38 
1.92 
2.57 

0.89 
1.02 
1.28 
1.53 
1.79 

2.04 
2.30 
2.55 
3.06 
3.57 

0.17 
0.22 
0.34 
0.47 
0.63 

0.57 
0.65 
0.82 
0.98 
1.14 

0.06 
0.08 
0.11 
0.16 
0.21 

0.45 
0.57 
0.68 
0.79 

0.03 
0.05 
0.07 
0.09 

80 
90 
100 
120 
140 

5.23 
5.88 
6.54 
7.84 
9.15 

7.9 
9.8 
12.0 
16.8 
22.3 

3.63 
4.09 
4.54 
5.45 
6.35 

3.28 
4.08 
4.96 
7.0 
9.2 

0.81 
1.0 
1.22 
1.71 
2.28 

1.31 
1.47 
1.63 
1.96 
2.29 

0.27 
0.34 
0.41 
0.58 
0.76 

0.91 
1.02 
1.13 
1.36 
1.58 

0.11 
0.14 
0.17 
0.24 
0.31 

160 
180 
200 
250 
300 

10.46 
11.76 
13.07 
16.34 
19.61 

29.0 
35.7 
43.1 
65.5 
92.0 

7.26 
8.17 
9.08 
11.35 
13.62 

11.8 
14.8 
17.8 
27.1 
38.0 

4.08 
4.60 
5.11 
6.38 
7.66 

2.91 
3.61 
4.4 
6.7 
9.3 

2.61 
2.94 
3.27 
4.08 
4.90 

0.98 
1.22 
1.48 
2.24 
3.14 

1.82 
2.05 
2.27 
2.84 
3.40 

0.41 
0.53 
0.61 
0.93 
1.29 

1.73 
2.21 
2.75 
3.35 
3.98 

350 
400 
450 
500 
550 

22.87 
26.14 
29.41 

122.0 
156.0 
196.0 

15.89 
18.16 
20.43 
22.70 
24.96 

50.5 
65.0 
81.0 
98.0 
117.0 

8.93 
10.21 
11.49 
12.77 
14.04 

12.4 
16.0 
19.8 
24.0 
28.7 

5.72 
6.54 
7.35 
8.17 
8.99 

4.19 
5.4 
6.7 
8.1 
9.6 

3.98 
4.54 
5.11 
5.68 
6.24 

600 
700 
800 
900 
1000 

27.23 

137.0 

15.32 
17.87 
20.42 
22.98 

33.7 
44.9 
57.0 
71.0 

9.80 
11.44 
13.07 
14.71 
16.34 

11.3 
15.1 
19.4 
24.0 
29.2 

6.81 
7.95 
.9.08 
10.22 
11.35 

4.68 
6.24 
7.98 
9.93 
12.04 

1100 
1200 
1300 
1400 
1500 

1 

17.97 
19.61 

34.9 
40.9 

12.49 
13.62 
14.  7« 
15.89 
!17.03 

14.4 
16.9 
19.6 
22.5 
25.6 

280 


ELEMENTS  OF  HYDRAULICS 


FRICTION  HEAD  IN  PIPES — (Continued) 


Discharge  in  thousands 
of  gallons  per  24  hourd 

0 

3 

S3 
0   6 

*  8 

I1 

•S  15 
.2  £ 
Q 

Inside  diameter  of  pipe 

8  in. 

10  in. 

12  in. 

16  in. 

20  in. 

Velocity  in  feet 
per  sec. 

Friction  head  in 
feet  per  100  ft. 

Velocity  in  feet 
per  sec. 

Friction  head  in 
feet  per  100  ft. 

Velocity  in  feet 
per  sec. 

Friction  head  in 
feet  per  100  ft. 

Velocity  in  feet 
per  sec. 

Friction  head  in 
feet  per  100  ft. 

Velocity  in  feet 
per  sec. 

a 
v£ 

II 

I* 

II 

200 
250 
300 
350 
400 

0.309 
0.386 
0.464 
0.541 
0.619 

0.89 
1.11 
1.33 
1.56 
1.77 

0.08 
0.12 
0.14 
0.22 
0.28 

0.57 
0.71 
0.85 
0.99 
1.13 

0.04 
0.05 
0.06 

0.07 
0.09 

0.39 
0.49 
0.59 
0.69 
0.79 

0.01 
0.02 
0.02 
0.03 
0.04 

0.22 
0.28 
0.33 
0.39 

0.44 

0.003 
0.004 
0.006 
0.008 
0.010 

0.28 

0.003 

450 
500 
550 
600 
700 

0.696 
0.774 
0.851 
0.928 
1.083 

1.99 
2.22 
2.44 
2.66 
3.10 

0.34 
0.42 
0.50 
0.59 

0.78 

1.28 
1.42 
1.56 
1.70 
1.99 

0.12 
0.14 
0.17 
0.20 
0.26 

0.89 
0.99 
.09 
.18 
.38 

0.05 
0.06 
0.07 
0.08 
0.11 

0.50 
0.55 
0.61 
0.66 
0.77 

0.012 
0.015 
0.017 
0.02 
0.03 

0.31 
0.35 
0.39 
0.43 
0.50 

0.004 
0.005 
0.006 
0.007 
0.009 

0.012 
0.014 
0.017 
0.020 
0.024 

800 
900 
1,000 
1,100 
1,200 

.238 
.392 
.547 
.702 
.857 

3.55 
3.99 
4.43 
4.88 
5.37 

0.99 
1.24 
1.51 
1.80 
2.11 

2.27 
2.55 
2.84 
3.12 
3.40 

0.34 
0.42 
0.51 
0.61 
0.71 

.58 
.77 
.97 
2.17 
2.36 

2.96 
3.94 
4.92 
5.91 
6.89 

0.14 
0.17 
0.21 
0.25 
0.29 

0.89 
1.00 
1.11 
1.22 
1.33 

0.03 
0.04 
0.05 
0.06 
0.07 

0.57 
0.64 
0.71 
0.78 
0.85 

1,500 
2,000 
2,500 
3,000 
3,500 

2.321 
3.094 
3.868 
4.642 
5.41 

6.65 
8.86 
11.08 
13.30 

3.18 
5.4 
8.4 
11.6 

4.26 
5.67 
7.10 
8.51 
9.93 

1.C8 
1.84 
2.78 
3.86 
5.19 

0.44 
0.76 
1.15 
1.60 
2.13 

1.66 
2.22 
2.77 
3.32 
3.88 

0.11 
0.19 
0.28 
0.40 
0.53 

1.06 
1.42 
1.77 
2.13 
2.48 

0.04 
0.06 
0.09 
0.13 
0.18 

4,000 
5,000 
6,000 
7,000 
8,000 

6.19 
7.74 
9.28 
10.83 
12.38 

11.35 
14.19 
17.03 

6.65 
10.05 
14.09 

7.88 
9.85 
11.82 
13.79 
15.76 

2.70 
4.10 
5.8 
7.7 
9.9 

4.43 
5.54 
6.65 

7.76 
8.86 

0.68 
1.02 
1.43 
1.90 
2.42 

2.84 
3.55 
4.26 
4.96 
5.67 

0.23 
0.34 

0.48 
0.64 
0.82 





9,000 
10,000 
11,000 
12,000 
15,000 

13.92 
15.47 
17.02 
18.57 
23.21 

17.73 
19.  7C 

12.2 
15.0 

9.97 
11.08 
12.19 
13.30 
16.62 

3.02 
3.68 
4.40 
5.2 

7.8 

6.38 
7.09 
7.80 
8.51 
10.64 

1.02 
1.24 
1.48 
1.74 
2.62 

16,000 
17,000 
18,000 
19,000 
20,000 

24.76 
26.30 
27.85 
29.40 
30.94 

11.35 
12.06 
12.77 
13.47 

14.18 

2.96 
3.31 
3.68 
4.07 
4.48 

TABLES 


281 


FRICTION  HEAD  IN  PIPES — (Continued) 


C    3 

0 

Inside  diameter  of  pipe 

0     0 

3  ,jg 

3 

3 

24  in. 

30  in. 

36  in. 

42  in. 

48  in. 

is 

a     . 

"S 

d  4$ 

(3 

o 

la 

•s«s 

£\ 

®    tn 

d    c5 

V     t, 

«£ 

.a 

li 

«2 

a 

li 

45 

.3 

|i 

* 
a 

li 

Jj 

C 

IS 

.32 

??    fl 

M    fe 

>>    o 

d  *•* 

>>  o 

rl      M 

>>  g 

§>_ 

>>  .: 

a  ** 

>>  ^ 

(3    *•" 

a  £ 
|1 

5  "s 

I: 
13 

li 

II 

1  a 

.s  -£ 

B| 

o  " 

|s 

Frictioi 
feet  pe 

•3 
,2  <3 
.ft 

~    ft 

11 

Velocit 
per  se( 

1! 

fj 

I 

»S. 

T^ 

1! 
li 

1  0 

1  5471  1  n  4Ql  n  on? 

0  32 

0  002 

0  22 

0  002 

1   5 

2  321 

0  74    0  015 

0  47 

0  005 

0  33 

0  003 

2  0 

3  094 

0  98    0  026 

0  63 

0  009 

0  44 

0  004 

2  5 

3  868 

1   23 

0  039 

0  79 

0  013 

0  55 

0  005 

3.0 

4.642 

1.48 

0.055 

0.95 

0.018 

0.66 

0.008 

0.48 

0.004 

3   5 

5  41 

1  72 

0  07 

1   10 

0  025 

0  77 

0  010 

0  56 

0  005 

4.0 

6.19 

1.97 

0.09 

1.26 

0.032 

0.88 

0.013 

0.64 

0.006 

0.49 

0.003 

4.5 

6.96 

2.22 

0.12 

1.42 

0.039 

0.99 

0.72 

0.007 

0.55 

0.004 

5.0 

7.74 

2.46 

0.14 

1.58 

0.048 

1.09 

0.020 

0.80 

0.009 

0.62 

0.005 

6.0 

9.28 

2.96 

0.20 

1.89 

0.067 

1.31 

0.027 

0.96 

0.013 

0.74 

0.007 

7.0 

10.83 

3.45 

0.26 

2.21 

0.09 

1.53 

0.036 

.13 

0.017 

0.86 

0.009 

8.0 

12.38 

3.94 

0.34 

2.52 

0.11 

1.75 

0.047 

.29 

0.022 

0.98 

0.012 

9.0 

13.92 

4.43 

0.42 

2.84 

0.14 

1.97 

0.058 

.45 

0.027 

.10 

0.014 

10.0 

15.47 

4.92 

0.51 

3.15 

0.17 

2.19 

0.071 

.61 

0.033 

.23 

0.017 

12.0 

18.57 

5.91 

0.71 

3.78 

0.24 

2.63 

0.099 

.93 

0.047 

.48 

0.024 

14.0 

21.66 

6.89 

0.95 

4.41 

0.32 

3.06 

0.13 

2.25 

0.06 

.72 

0.032 

16.0 

24.76 

7.88 

1.22 

5.04 

0.41 

3.50 

0.17 

2.57 

0.08 

.97 

0.042 

18.0 

27.85 

8.86 

1.52 

5.67 

0.51 

3.94 

0.21 

2.89 

0.10 

2.22 

0.052 

20.0 

30.94 

9.85 

1.83 

6.30 

0.62 

4.38 

0.25 

3.22 

0.12 

2.46 

0.063 

22.0 

34.04 

10.83 

2.19 

6.93 

0.74 

4.82 

0.30 

3.53 

0.14 

2.71 

0.075 

24.0 

37.13 

11.82 

2.59 

7.56 

0.87 

5.25 

0.36 

3.86 

0.17 

2.96 

0.09 

26.0 

40.23 

12.80 

2.99 

8.20 

1.01 

5.69 

0.41 

4.18 

0.20 

3.20 

0.10 

28.0 

43.32 

13.79 

3.42 

8.83 

1.16 

6.13 

0.48 

4.50 

0.22 

3.45 

0.12 

30.0 

46.42 

14.77 

3.90 

9.46 

1.32 

6.57 

0.54 

4.82 

0.26 

3.69 

0.13 

32.0 

49.51 

10.09 

1.48 

7.00 

0.61 

5.15 

0.29 

3.94 

0.15 

34.0 

52.6 

10.72 

1.66 

7.44 

0.68 

5.47 

0.32 

4.19 

0.17 

36.0 

55.7 

11.35 

1.84 

7.88 

0.76 

5.79 

0.36 

4.43 

0.19 

38.0 

58.8 

11.98 

2.04 

8.32 

0.84 

6.11 

0.40 

4.68 

0.21 

40.0 

61.9 

8.76 

0.92 

6.45 

0.44 

4.92 

0.23 

50.0 

77.4 

10.95 

1.39 

8.04 

0.66 

6.16 

0.34 

60.0 

92.8 

13.13 

1.96 

9.65 

0.92 

7.39 

0.48 

70.0 

108.3 

11.26 

1.22 

8.62 

0.64 

80.0 

123.8 

12.86 

1.57 

9.85 

0.82 

90.0 

139.2 

11.08 

1.02 

100.0 

154.7 







12.31 

1.24  ' 

282 


ELEMENTS  OF  HYDRAULICS 


TABLE  14. — BAZIN'S   VALUES  OF  CHEZY'S  COEFFICIENT 

Values  of  the  coefficient  C  in  Chezy's  formula  v  =  C\/Vs  according  to  Bazin's  formula 
(Art.  24): 

C  — 


87 


m 
0.552  +  - 

Vr 


Coefficient  of  roughness,  m 

Hydraulic 
radius 
r,  in 
feet 

Planed  tim- 
ber or 
smooth 
cement 

Unplaned 
timber,  well 
laid  brick, 
or  concrete 

Ashlar,  good 
rubble  mas- 
onry,    or 
poor   brick- 
work 

Earth  in 
good 
condition 

Earth  in 
ordinary 
condition 

Earth  in 
bad 
condition 

TO  =  0.06 

TO  =  0.16 

TO  =  0.46 

TO  =  0.85 

TO  =  1.30 

TO  =  1.75 

0.1 

117 

82 

43 

27 

19 

14 

0.2 

127 

96 

55 

35 

25 

19 

0.3 

131 

103 

63 

41 

30 

23 

0.4 

135 

108 

68 

46 

33 

26 

0.5 

136 

112 

71 

50 

3a 

29 

0.6 

138 

115 

.       76 

53 

39 

31 

0.7 

139 

117 

79 

55 

41 

33 

0.8 

141 

119 

82 

58 

43 

35 

0.9 

141 

121 

84 

60 

45 

36 

1.0 

142 

122 

86 

62 

47 

38 

1.25 

143 

125 

90 

66 

51 

41 

1.50 

145 

127 

94 

70 

54 

44 

1.75 

145 

129 

97 

73 

57 

47 

2.00 

146 

131 

99 

75 

59 

49 

2.5 

147 

133 

104 

80 

63 

53 

3.0 

.      148 

135 

106 

83 

67 

57 

4.0 

150 

138 

111 

89 

72 

61 

5.0 

150 

140 

115 

93 

77 

65 

6.0 

151 

141 

118 

97 

80 

69 

7.0 

152 

142 

120 

100 

83 

72 

8.0 

152 

143 

122 

102 

86 

74 

9.0 

152 

144 

123 

104 

88 

77 

10.0 

152 

145 

125 

106 

90 

79 

12.0 

153 

145 

127 

109 

94 

82 

15.0 

153 

147 

130 

113 

98 

86 

20.0 

154 

148 

133 

117 

103 

92 

30.0 

155 

150 

137 

123 

110 

100 

40.0 

155 

151 

139 

127 

115 

105 

50.0 

155 

151 

141 

129 

118 

109 

TABLES 


283 


TABLE  15.  —  KUTTER'S  VALUES  OF  CHEZY'S  COEFFICIENT 


Values  of  the  coefficient  C  in  Chezy's  formula  v 
(Eq.  (67),  Art.  24): 

41.  65  + 


according  to  Kutter's  formula 


Slope, 

Coefficient  of 

Hydraulic  radius  r,  in  feet 

s       roughness,  n 

0.1  0.20.4 

0.6|0.8 

1  |l.5|  2  |  3  |  4  |  6  |  8 

10 

15  20 

0.009 

65 

87 

111J127 

138  148 

166  179 

197  209  226t238,246  262J271 

<o 

0.010 

57 

75 

97 

112 

122  131 

148  160 

177  188  206  216  225  240  249 

1 

•   0.011 

50 

67 

87 

100 

109  118 

133 

144 

160  172  188  199  207  222  231 

b 

0.012 

44 

59 

78 

90 

99  106 

121 

131 

147  158  174  184 

192  206 

215 

0) 

o  a 

0.013 

40 

53 

70 

81 

90 

97 

111 

121 

135146 

161 

171 

179 

193 

202 

O  _,  00 

0.017 

28 

38 

51 

60 

66 

72 

83 

91 

103  113 

126 

135 

142 

155  164 

O  fl  rH 

0.020 

23 

31 

42 

49 

55 

60 

69 

77 

88  96 

108 

117 

124 

136 

1144 

d  r-i  d 

0.025 

17 

24 

32 

38 

43 

47 

55 

61 

70 

78 

•  88 

96 

102 

114 

121 

II  II  II 

0.030 

14 

19 

26 

31 

35 

38 

45 

50 

59  65 

74 

82 

87 

98106 

00 

0.035 

12 

16 

22 

26 

30 

32 

38 

43 

50  56 

64 

71 

76 

86 

94 

0.009 

78  100 

124 

139 

150 

158 

173  184 

198  207 

220 

228 

234 

244 

250 

0 

0.010 

67 

87 

109 

122 

133 

140 

154 

164 

178 

187 

199 

206 

212 

220  228 

•"5 

0.011 

59 

77 

97 

109 

119 

126 

139 

148 

161 

170 

182 

189 

195 

205211 

£ 

0.012 

52 

68 

88 

98 

107 

114 

126 

135 

148 

156 

168 

175 

181 

189 

196 

0.013 

47 

62 

79 

90 

98 

104 

116 

124 

136 

145 

156 

163 

169 

179 

184 

|  g  | 

0.017 

33 

44 

57 

65 

71 

77 

87 

94 

104 

111 

122 

129 

134 

142 

149 

O  .g  <N 

0.020 

26 

35 

46 

53 

59 

64 

72 

79 

88 

95 

105 

111 

116 

125 

131 

d  ^  d 

0.025 

20 

26 

35 

41 

46 

49 

57 

62 

71 

77 

85 

91 

96 

104 

110 

II  II  II 

0.030 

10 

21 

28 

33 

37 

40 

47 

51 

59 

64 

72 

78 

82 

90 

96 

=C 

0.035 

13 

18 

24 

28 

31 

34 

40 

44 

50 

56 

63 

68 

72 

85 

0.009 

90 

112 

136 

149  158 

166 

178 

187 

198 

206 

215 

221 

226 

233 

237 

5« 

0.010 

78 

98 

119 

131  140 

147 

159 

168 

178 

186 

195 

201 

205 

212216 

s 

0.011 

68 

86 

106 

118 

126 

132  144 

151 

162 

169 

178 

184 

188 

195  200 

3 

O  ft 

0.012 

00 

76 

95 

105 

114 

120 

130 

138 

149 

155 

164 

170 

174 

181 

185 

o  ^ 

0.013 

54 

69 

86 

96 

103 

109 

120 

127 

137 

143 

152 

158 

162 

169 

173 

rl  O"  "* 

O  r-i  00 

o  .5  S 

0.017 

37 

48 

62 

70 

76 

81 

89 

96 

104 

111 

119 

124 

128 

135 

139 

0  rH  0 

0.020 

30 

39 

50 

57 

63 

67 

75 

81 

89 

94 

102 

107 

111 

118 

122 

II  II  II 

0.025 

22 

29 

38 

44 

48 

52 

59 

64 

71 

76 

84 

88 

92 

98 

102 

CO 

0.030 

17 

23 

31 

35 

39 

42 

48 

53 

59 

64 

71 

75 

78 

85 

89 

0.035 

14 

19 

25 

30 

33 

35 

41 

45 

51 

55 

61 

66 

69 

75 

79 

284 


ELEMENTS  OF  HYDRAULICS 


KUTTER'S  VALUE  OF  CHEZY'S  COEFFICIENT — (Continued) 


Slope, 

Coefficient  of 

Hydraulic  radius  r  in  feet 

s 

roughness,  -n 

0.10.2|0.3J0.4j0.60.8  1.0  1.5  2 

3   4   6   10  15 

20 

o 

0.009 
0.010 

99 
85 

121 
105 

133 
116 

143 
125 

155 

138 

164 
145 

170 
151 

181 
162 

188 
170 

200 
179 

205 
185 

213  222 
193  201 

228231 
207  210 

g 

0.011 

74 

93 

103 

112 

122 

131 

136 

146 

154 

163 

168 

176 

185 

190 

194 

<3 

0.012 

65 

83 

92 

100 

111 

118 

123 

133 

140 

149 

155 

162 

170 

176 

180 

ft 
0  ^ 

0.013 

59 

74 

83 

91 

100 

107 

113 

122 

129 

137 

143 

150 

158 

164 

168 

IM  O  "*" 

0.017 

41 

52 

59 

65 

73 

79 

83 

91 

97 

105 

111 

117 

125 

131 

134 

°.  —<  °. 

0.020 

32 

42 

48 

53 

60 

65 

69 

77 

82 

89 

94 

100 

108 

113 

117 

O  >-H  --" 
II   II   II 

0.025 

24 

31 

36 

40 

46 

50 

54 

60 

64 

72 

76 

82 

89 

95 

98 

00 

0.030 

18 

25 

29 

32 

37 

41 

44 

49 

54 

59 

63 

69 

76 

82 

85 

0.035 

15 

21 

24 

27 

31 

34 

37 

42 

45 

51 

55 

60 

67 

72 

76 

0.009 

104 

126 

138 

148 

157 

166 

172 

183 

190 

199 

204 

211 

219 

224 

227 

JH 

0.010 

89 

110 

120 

129 

140 

148 

154 

164 

170 

179 

184 

191 

199  203 

207 

H 

0.011 

78 

97 

107 

115 

126 

133 

138 

148 

154 

162 

168 

175 

183 

187 

190 

S 

0.012 

69 

87 

96 

104 

113 

121 

125 

135 

141 

149 

154 

161 

168 

172 

176 

0  ^ 

0.013 

62 

78 

87 

94 

103 

110 

115 

124 

130 

138 

142 

149 

157 

162 

164 

3f  35 

~ 

°  fl  2 

0.017 

43 

54 

62 

68 

75 

81 

85 

93 

98 

105 

110 

116 

123 

128 

131 

°.  •"  ""! 

0.020 

34 

44 

50 

55 

62 

67 

70 

78 

83 

89 

94 

99 

107 

110 

115 

ii  ii  ii 

0.025 

25 

32 

37 

42 

47 

51 

55 

61 

65 

71 

76 

81 

88 

92 

96 

00 

0.030 

19 

25 

30 

33 

38 

42 

45 

50 

54 

59 

63 

69 

75 

80 

83 

0.035 

16 

21 

24 

27 

31 

35 

37 

42 

45 

51 

55 

60 

66 

70 

73 

0.009 

110 

129 

141 

150 

161 

169 

175 

184 

191 

199 

204 

211 

218 

222 

225 

0.010 

94 

113 

124 

1'31 

142 

150 

155 

165 

171  179 

184 

190 

197 

202205 

jg 

0.011 

83 

99 

109 

117 

127 

134 

139 

149 

155 

163 

168 

174 

181 

186 

188 

b 

0.012 

73 

89 

98 

105 

115 

122 

127 

136 

142 

149 

154 

160 

167 

171 

175 

0  & 

0.013 

65 

81 

89 

96 

104 

111 

116 

124 

130 

138 

142 

149 

155 

160 

163 

§c  22 

0.017 

45 

57 

63 

69 

76 

82 

86 

93 

98 

105 

110 

116 

122 

127 

129 

0.020 

36 

45 

51 

56 

63 

68 

71 

78 

83 

89 

93 

99 

105 

110 

113 

II  II  II 

0.025 

27 

34 

39 

43 

48 

52 

56 

62 

66 

71 

75 

81 

87 

91 

94 

0.030 

21 

27 

30 

34 

39 

42 

45 

50 

54 

59 

63 

68 

74 

78 

81 

0.035 

17 

22 

25 

28 

32 

35 

38 

43 

46 

51 

54 

59 

65 

68 

72 

0.009 

110 

130 

143 

151 

162 

170 

175 

185 

191 

199 

204 

210 

217 

222 

225 

0.010 

95 

114 

125 

133 

143 

151 

156 

165 

171 

179 

184 

190 

196 

200 

204 

J! 

0.011 

83 

100 

111 

119 

129 

135 

141 

149 

155 

162 

167 

173 

180 

184 

187 

S 

0.012 

74 

90 

100 

107 

116 

123 

128 

136 

142 

149 

154 

160 

166 

170 

173 

8. 

0.013 

66 

81 

90 

98 

106 

112 

117 

125 

130 

138 

142 

148 

154 

159 

161 

,-H  "~ 

rH      00 

0.017 

46 

57 

64 

70 

77 

82 

87 

94 

99 

105 

109 

115 

121 

126 

128 

o  .g  • 

0.020 

36 

46 

52 

57 

64 

68 

72 

79 

83 

89 

93 

99 

105 

108 

112 

II  II  II 

0.025 

27 

34 

39 

44 

49 

53 

56 

62 

66 

71 

76 

81 

86 

90 

93 

00 

0.030 

21 

27 

31 

35 

39 

43 

45 

51 

55 

59 

63 

68 

74 

77 

80 

0.035 

17 

22 

25 

29 

33 

35 

38 

43  1  46 

51 

55 

59 

65 

68|  71 

TABLES 


285 


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286 


ELEMENTS  OF  HYDRAULICS 


TABLE  17. — DISCHARGE  PER  INCH  OF  LENGTH  OVER  RECTANGULAR  NOTCH 

WEIRS 

Discharge  over  sharp-crested,  vertical,  rectangular  notch  weirs  in  cubic  feet  per  minute 

per  inch  of  length. 
Computed  from  Eq.  (30),  Art.  10:  Q  =  0.4  bh^  for  b  =  1  in. 


Depth  on  crest  in 
inches 

0 

1/8 

1/4 

3/8 

1/2 

5/8 

3/4 

7/8 

0 

0.00 

0.01 

0.05 

0.09 

0.14 

0.19 

0.26 

0.32 

1 

0.40 

0.47 

0.55 

0.64 

0.73 

0.82 

0.92 

1.02 

2 

1.13 

1.23 

1.35 

1.46 

1.58 

1.70 

1.82 

1.95 

3 

2.07 

2.21 

2.34 

2.48 

2.61 

2.76 

2.90 

3.05 

4 

3.20 

3.35 

3.50 

3.66 

3.81 

3.97 

4.14 

4.30 

5 

4.47 

4.64 

4.81 

4.98 

5.15 

5.33 

5.51 

5.69 

6 

5.87 

6.06 

6.25 

6.44 

6.62 

6.82 

7.01 

7.21 

7 

7.40 

7.60 

7.80 

8.01 

8.21 

8.42 

8.63 

8.83 

8 

9.05 

9.26 

9.47 

9.69 

9.91 

10.13 

10.35 

10.57 

9 

10.80 

11.02 

11.25 

11.48 

11.71 

11.94 

12.17 

12.41 

10 

12.64 

12.88 

13.12 

13.36 

13.60 

13.85 

14.09 

14.34 

11 

14.59 

14.84 

15.09 

15.34 

15.59 

15.85 

16.11 

16.36 

12 

16.62 

16.88 

17.15 

17.41 

17.67 

17.94 

18.21 

18.47 

13 

18.74 

19.01' 

19.29 

19.56 

19.84 

20.11 

20.39 

20.67 

14 

20.95 

21.23 

21.51 

21.80 

22.08 

22.37 

22.65 

22.94 

15 

23.23 

23.52 

23.82 

24.11 

24.40 

24.70 

25.00 

25.30 

16 

25.60 

25.90 

26.20 

26.50 

26.80 

27.11 

27.42 

27.72 

17 

28.03 

28.34 

28.65 

28.97 

29.28 

29.59 

29.91 

30.22 

18 

30.54 

30.86 

31.18 

31.50 

31.82 

32.15 

32.47 

32.80 

19 

33.12 

33.45 

33.78 

34.11 

34.44 

34.77 

35.10 

35.44 

20 

35.77 

36.11 

36.45 

36.78 

37.12 

37.46 

37.80 

38.15 

21 

38.49 

38.84 

39.18 

39.53 

39.87 

40.24 

40.60 

40.96 

22 

41.28 

41.64 

41.98 

42.36 

42.68 

43.04 

43.44 

43.76 

23 

44.12 

44.48 

44.84 

45.20 

45.56 

45.96 

46.32 

46.68 

24 

47.04 

47.40 

47.76 

48.12 

48.52 

48.88 

49.28 

49.64 

25 

50.00 

50.40 

50.76 

51.08 

51.52 

51.88 

52.28 

52.64 

26 

53.04 

53.40 

53.80 

54.16 

54.56 

54.96 

55.36 

55.72 

27 

56.12 

56.52 

56.92 

57.32 

57.68 

58.08 

58.48 

58.88 

28 

59.28 

59.68 

60.08 

60.48 

60.84 

61.28 

61.68 

62.08 

29 

62.48 

62.88 

63.28 

63.68 

64.08 

64.52 

64.92 

65.32 

30 

65  .  72 

66.16 

66.56 

66.96 

67.36 

67.80 

68.20 

68.64 

TABLES 


287 


TABLE  18. — DISCHARGE  PER  FOOT  OF  LENGTH  OVER  RECTANGULAR  NOTCH 

WEIRS 

Discharge  over  sharp  crested,  vertical,  rectangular  notch   weirs  in  cubic  feet  per  second 
per  foot  of  length.     Computed  from  Eq.  (29),  Art.  10: 

Q  =  3. 3bh 3/2  for  6  =  1  ft. 


Depth  on  crest 
in  feat 

0.00 

0.01 

0.02 

0.03 

0.04 

0.05 

0.06 

0.07 

0.08 

0.09 

0.0 

0.000 

0.003 

0.009 

0.017 

0.026 

0.037 

0.049 

0.061 

0.075 

0.089 

0.1 

0.104 

0.120 

0.137 

0.155 

0.173 

0.192 

0.211 

0.231 

0.252 

0.273 

0.2 

0.295 

0.317 

0.341 

0.364 

0.388 

0.413 

0.438 

0.463 

0.489 

0.515 

0.3 

0.542 

0.570 

0.597 

0.626 

0.654 

0.683 

0.713 

0.743 

0.773 

0.804 

0.4 

0.835 

0.866 

0.898 

0.931 

0.963 

0.996 

1.030 

1.063 

1.098 
1.458 

1.132 

0.5 

1.167 

1.202 

1.238 

1.273 

1.309 

1.346 

1.383 

1.420 

1.496 

0.6 

1.538 

1.572 

.  1.611 

1.650 

1.690 

1.729 

1.769 

1.810 

1.850 

1.892 

0.7 

1.933 

1.974 

2.016 

2.058 

2.101 

2.143 

2.187 

2.230 

2.273 

2.317 

0.8 

2.361 

2.406 

2.450 

2.495 

2.541 

2.586 

2.632 

2.678 

2.724 

2.768 

0.9 

2.818 

2.865 

2.912 

2.960 

3.008 

3.055 

3.104 

3.152 

3.202 

3.251 

1.0 

3.300 

3.350 

3.399 

3.449 

3.501 

3.551 

3.600 

3.653 

3.703 

3.755 

1.1 

3.808 

3.858 

3.911 

3.963 

4.016 

4.069 

4.122 

4.178 

4.231 

4.283 

1.2 

4.340 

4.392 

4.448 

4.501 

4.557 

4.613 

4.666 

4.722 

4.778 

4.834 

1.3 

4.891 

4.947 

5.006 

5.062 

5.118 

5.178 

5.234 

5.293 

5.349 

5.409 

1.4 

5.468 

5.524 

5.584 

5.643 

5.702 

5.762 

5.82'l 

5.881 

5.940 

6.003 

1.5 

6.062 

6.125 

6.184 

6.247 

6.306 

6.369 

6.428 

6.491 

6.554 

6.617 

1.6 

6.679 

6.742 

6.805 

6.867 

6.930 

6.993 

7.059 

7.121 

7.187 

7.250 

1.7 

7.316 

7.379 

7.445 

7.508 

7.573 

7.639 

7.706 

7.772 

7.838 

7.904 

1.8 

7.970 

8.036 

8.102 

8.171 

8.237 

8.303 

8.372 

8.438 

8.507 

8.573 

1.9 

8.643 

8.712 

8.778 

8.847 

8.917 

8.986 

9.055 

9.125 

9.194 

9.263 

2.0 

9.332 

9.405 

9.474 

9.544 

9.616 

9.686 

9.758 

9.827 

9.900 

9.969 

2.1              10.042 
2.2             10.768 
2.3             11.510 

10.115 

10.841 
11.586 

10.187 
10.916 
11.662 

10.260 
10.989 
11.738 

10.332 
11.065 
11.814 

10.405 
11.138 
11.887 

10.478 
11.213 
11.966 

10.550 
11.286 
12.042 

10.623 
11.362 
12.118 

10.695 
11.435 
12.194 

2.4 

12.269 

12.345 

12.425 

12.500 

12.576 

12.656 

12.731 

12.811 

12.890 

12.966 

2.5 

12.935 

13.124 

13  .  200 

13.279 

13.358 

13.438 

13.517 

13.596 

13.675 

13  .  754 

2.6 

2.7 

13  .  834 
14.642 

13.916 
14.721 

13.995 
14  .  804 

14.075 

14.886 

14.154 
14.969 

14.236 
15.048 

14.315 
15.131 

14.398 
15.213 

14.477 
15.296 

14  .  560 
15.378 

2.8 

15.461 

15.543 

15.629 

15.711 

15.794 

15.876 

15.962 

16.045 

16.130 

16.213 

2.9 

16.299 

16.381 

16.467 

16.550 

16.635 

16.721 

16.807 

16.889 

16.975 

17.061 

3.0 

17.147 

17.233 

17.318 

17.404 

17.490 

17.579 

17.665 

17.751 

17.837 

17.926 

3.1 

18.011 

18.101 

18.186 

18.275 

18.361 

18.450 

18.536 

18.625 

18.714 

18.803 

3.2 

18.889 

18.978 

19.067 

19.157 

19  .  246 

19.335 

19.424 

19.513 

19  .  602 

19.694 

3.3 

19.784 

19.873 

19.962 

20.054 

20.143 

20.236 

20.325 

20.414 

20.506 

20.599 

3.4 

20.688 

20.780 

20.873 

20.962 

21.054 

21.146 

21.239 

21.331 

21.424 

21.516 
22.447 

3.5 

21  .  608 

21.701 

21.793 

21.886 

21.978 

22.074 

22.166 

22.259 

22.354 

3.6 

22.542 

22.635 

22.730 

22.823 

22.919 

23.011 

23.107 

23  .  202 

23.295 

23.390 

3.7 

23.486 

23.582 

23.678 

23.773 

23.869 

23.965 

24.060 

24.156 

24.252 

24.347 

3.8 

24.446 

24.542 

24.638 

24.734 

24.833 

24.928 

25.027 

25.123 

25.222 

25.318 

3.9 

25.417 

25.516 

25.611 
26.598 

25.710 

25.809 

25.905 

26.004 

26.103 

26-.  202 

26.301 

4.0 

26.400 

26.499 

26.697 

26.796 

26.895 

26.997 

27.096 

27.  195 

27.298 

288 


ELEMENTS  OF  HYDRAULICS 


TABLE  19.— PRINCIPLES 


Kinematics  (motion) 

Linear  motion 

Angular  motion 

s  =  displacement 

6  =  displacement 

.  v  =  velocity 

w  =  velocity 

a  =  acceleration 

a  =  acceleration 

vo  =  initial  velocity 

coo  =  initial  velocity 

F  =  force 

M  =  torque  about  fixed  axis 

Notation 

W  =  Fs  =  work 

W  =  MB  =  work 

<n  =  mass 

/  =  2mr2  =  moment  of  inertia 

w  =  mg  =  weight 

t  =  time 

t  =  time 

Mt  =  impulse 

Ft  =  impulse 

Jco  =  momentum 

mv  =  momentum 

Definitions 

da           dv        d*s 

dO             da        d*6 
~  dt'   a  ~  dt    ~  dt" 

Uniformly 

v  =  vo  +  at 

co  =   coo  +  at 

accelerated  motion 

s  =  vot  +  %at2 

0  =   coo<  +  Ja<2 

accel.  =  const. 

W2  =  V02  -|-  2as 

co.2  -   w°2  +  2«e 

d2s              ds 

dzO              dO 

dT*    =  Of  dt    =  at  +  C 

t-                     "TV  =  a>  ~v~  —  at  -\-  Ci 

Derivation  of 
above  formulas 

s  =  ia<2  +  Cit  +  Cz 

if  t  =  o,  v  =  uo  :.Ci  = 

vo                  If  t  =  0,  co  =  coo  .'.Ci  =   coo 

If  t  =  0,  s  =  0   .'.C2  = 

o            if  t  =  o,  e  =  o  .'.C2  =  o 

Relation  between 

v  =  rco 

linear  and 

at  =  ra 

at  —  tang.  comp.  of  accel. 

angular  motion 

an  =  v2/r  =  rco2 

an  =  normal  comp.  of  accel. 

v  7? 

Arc  AB  —  ds  =  rdO               v  =  rco 

Derivation  of  first 

.\ 

ds            dO                dv            dta 

two  formulas 

T/'     \ds 

dt   ~   T  dt                 d?    =  r  dT 

£..-  \A. 

v  —  rco                          at  =  ra 

A  v  B 

If  body  at  A  were  free,  it  would  proceed 

in    direction    of   tangent    AB   and   in 

time  t  would  reach  B  where  AB  =  vt. 

Derivation  of 

Since  it  is  found  at  C  instead  of  B  it 

normal  accel. 

must     have     experienced     a     central 

for  uniform 
circular  motion 

acceleration. 
Let  on     denote     this     central     acceleration.     Then     BC   =    %ant2- 
By  geometry  BC  X  BD  =  A  B2  and  in  the  limit  BD  approaches  2r. 

Hence  |ani2  X  2r  =  u2i2, 

from  which  an  —  v2/r  =  co2r. 

TABLES 


289 


OF  MECHANICS 


Dynamics  (force) 

|                   Linear  motion                                     Angular  motion 

Fundamental  law 

F  =  ma 

M  =  la 

Discussion  and 
derivation 

By  experiment  it  is  found  that 
F    ex    a     (Newton's    2nd  Law) 
.'.  F/a  =  const.,  say  m,  whence 
F  =  ma.      m  =  intrinsic     prop- 
erty  of   body   called  its   mass. 
Mass  =  measured  inertia. 

Consider    rota- 
Fk.         tion    of    rigid 
O         r                   body  about  a 

\m      fixed      axis. 
1          Then  for  a  par- 
ticle   of   mass 
m  at  distance  r  from  axis  of  rota- 
tion, law  F  =   ma  becomes  Fr  = 
mar,     or    since    a  =  ra,    Fr  = 
mr2a. 
By  summation 
2Fr  =  Smr2a 
But    ZFr  =  M  ,    and    Smr2  =  I 
.  .M  =  la. 

If  F  =  0  then  a  =  0  and  hence 
v  =  0    or    constant,    which    ex- 
presses Newton's  1st  Law. 

F  =  impressed  force,  ma  =  kine- 
tic   reaction    or    inertia   force. 
Equality  F  —  ma   is   dynamical 
expression  of  Newton's  3rd  Law. 

Principle  of  work 
and  energy 

w  =  Fs  =  m£_m£ 

„.„,.!£-!£ 

Derivation 

F  =  TOO,  »2  =  t>o2  +  2as 

1,2     -    VO* 

M  =  la,  w2  =  wo2  +  2a0 
a,2  -  wo2 

a 

TOU2            TWO2 

TT         MO        I    0        J"2    "  /W°2 

W  -  Fs  -  mas  -     2            2 

Principle  of  impulse 
and  momentum 

Fl  =  mv  —  mvo 

Mt  =  Iw  -  Iwo 

Derivation 

F  =  ma,  v  =  vo  +  at 
.'.  at  =  v  —  vo,  and 
Ft  —  mat  =  mv  —  mvo 

M  =  la,    u  =  wo  ~\-  at 
.'  .  at  =  w  —  wo,  and 
Mt  =  lat  =  lu  —  luo 

Power 

Fv 
Power  =  Fv,  h.p.  =  ^—7: 

Mw 
Power  =  Mct>,  h.p.  =  ^7: 

OoU 

Centrifugal  force 

w             w   v2        w 
F,  =  -  an  =  -   —  =  —  w2r 
c       g            g    r       g 

Derivation 

w                    v^ 
Special  case  of  F  =  ma  where  m  =  —  and  a  =  — 

D'Alembert's 
principle 

,—£-0 

Explanation  and 
use 

F  =  ma  where  F  =  external  impressed  force  and  a  =  accel.  produced. 
Introduce  another  force  P,  given  by  P  =   —  ma.  Then  by  addition, 
F  +  P  =  0;  i.e.,  the  body  is  in  equilibrium  under  the  action  of  F  and 
P.   P  is  called  the  kinetic  reaction,  or  reversed  effective  force,  since 
P  =  —  F.   By  introducing  this  idea  of  the  kinetic  reactions  equili- 
brating the  impressed  forces,  all  problems  in  dynamics  are    reduced 
to  statical  problems.     This  is  called  d'Alembert's  Principle,  and  is 
usually  expressed  in  the  form 

,-,g-0. 

290 


ELEMENTS  OF  HYDRAULICS 


TABLE  20. l — DISCHARGE  PER  FOOT  OP  LENGTH  OVER  SUPPRESSED  WEIRS 

Discharge  over  sharp-crested,  vertical,  suppressed  weirs  in  cubic  feet  per  second  per  foot 
of  length.     Computed  by  Bazin's  formula  (Art.  10) : 


Q  =  (0.405 


0 . 00984 


)(l  +0.55 


bhVZffh,  for  b  =  1  ft. 


Head  on  crest,  h, 

Height  of  weir,  d,  in  feet 

in  feet 

2 

4               6               8              10      |      20 

30 

0.1 

0.13 

0.13 

0.13 

0.13 

0.13 

0.13 

0.13 

0.2 

0.33 

0.33 

0.33 

0.33 

0.33 

0.33 

0.33 

0.3 

0.58 

0.58 

0.58 

0.58 

0.58 

0.58 

0.58 

0.4 

0.88 

0.88 

0.87 

0.87 

0.87 

0.87 

0.87 

0.5 

1.23 

1.21 

1.21 

1.21 

1.21 

1.20 

1.20 

0.6 

1.62 

1.59 

1.58 

1.58 

1.57 

1.57 

1.57 

0.7 

2.04 

1.99 

1.98 

1.98 

1.97 

1.97 

1.97 

0.8 

2.50 

2.43 

2.41 

2.41 

2.40 

2.40 

2.40 

0.9 

3.00 

2.90 

2.88 

2.86 

2.86 

2.85 

2.85 

1.0 

3.53 

3.40 

3.36 

3.35 

3.34 

3.33 

3.33 

.1 

4.09 

3.92 

3.87 

3.86 

3.85 

3.84 

3.84 

.2 

4.68 

4.48 

4.42 

4.40 

4.38 

4.36 

4.36 

.3 

5.31 

5.07 

4.99 

4.96 

4.94 

4.91 

4.91 

.4 

5.99 

5.68 

5.58 

5.54 

5.52 

5.49 

5.48 

.5 

6.68 

6.30 

6.20 

6.16 

6.13 

6.10 

6.09 

.6 

7.40 

6.97 

6.84 

6.78 

6.74 

6.69 

6.69 

.7 

8.14 

7.66 

7.49 

7.42 

7.39 

7.33 

7.32 

.8 

8.93 

8.37 

8.18 

8.09 

8.05 

7.98 

7.96 

1.9 

9.75 

9.11 

8.89 

8.79 

8.74 

8.65 

8.63 

2.0 

10.58 

9.87 

9.62 

9.51 

9.44 

9.34 

9.32 

2.1 

11.45 

10.65 

10.37 

10.25 

10.17 

10.05 

10.02 

2.2 

12.34 

11.46 

11.14 

10.99 

10.91 

10.78 

10.75 

2.3 

13.24 

12.29 

11.93 

11.77 

11.66 

11.52 

11.48 

2.4 

14.20 

13.15 

12.75 

12.56 

12.45 

12.28 

12.24 

2.5 

15.17 

14.03 

13.59 

13.38 

13.26 

13.06 

13.01 

2.6 

16.16 

14.92 

14.44 

14.20 

14.07 

13.85 

13.80 

2.7 

17.18 

15.83 

15.31 

15.04 

14.92 

14.65 

14.60 

2.8 

18.23 

16.79 

16.21 

15.92 

15.76 

15.48 

15.42 

2.9 

19.29 

17.77 

17.11 

16.79 

16.63 

16.33 

16.25 

3.0 

20.39 

18.74 

18.06 

17.71 

17.52 

17.18 

17.10 

3.1 

21.50 

19.74 

19.02 

18.64 

18.42 

18.04 

17.96 

3.2 

22.64 

20.77 

19.98 

19.58 

19.34 

18.93 

18.83 

3.3 

23.81 

21.80 

20.98 

20.55 

20.27 

19.82 

19.73 

3.4 

24.98 

22.89 

21.99 

21.52 

21.24 

20.75 

20.63 

3.5 

26.20 

24.00 

23.01 

22.48 

22.22 

21.69 

21.60 

3.6 

27.41 

25.09 

24.06 

23.52 

23.20 

22.62 

22.48 

3.7 

28.64 

26.22 

25.14 

24.56 

24.20 

23.59 

23.43 

3.8 

29.94 

27.38 

26.22 

25.60 

25.23 

24.56 

24.39 

3.9 

31.21 

28.53 

27.33 

26.65 

26.26 

25.53 

25.34 

4.0 

32.54 

29.74 

28.45 

27.74 

27.32 

26.55 

26.35 

4.1 

33.85 

30.95 

29.59 

28.83 

28.36 

27.55 

27.33 

4.2 

35.22 

32.18 

30.75 

29.96 

29.48 

28.59 

28.36 

4.3 

36.59 

33.43 

31.93 

31.10 

30.58 

29.62 

29.37 

4.4 

37.99 

34.70 

33.12 

32.24 

31.70 

30.66 

30.42 

4.5 

39.40 

35.98 

34.33 

33.39 

32.83 

31.74 

31.47 

4.6 

40.83 

37.29 

35.56 

34.58 

33.98 

32.84 

32.53 

4.7 

42.29 

38.62 

36.82 

35.75 

35.13 

33.93 

33.61 

4.8 

43.75 

39.96 

38.07 

37.00 

36.33 

35.05 

34.70 

4.9 

45.22 

41.30 

39.35 

38.20 

37.49 

36.15 

35.77 

5.0 

46.71 

42.67 

40.62 

39.44 

38.70 

37.28 

36.88 

Compiled  from  extensive  hydraulic  tables  by  Williams  and  Hazen. 


INDEX 


Accumulator,  hydraulic,  9 
Adjutage,  Venturi,  63 
American  type  of  turbine,  170 
Aqua  Claudia,  137 
Aqueducts,  comparison  of,  136 
Archimedes,  theorem  of,  21 

B 

Backwater,  118 
Barge  canal,  N.  Y.  State,  96 
Barker's  mill,  155 
Barometer,  mercury,  18 

water,  17 
Bazin's  coefficient,  280 

formula,  102 

Bends  and  elbows,  head  lost  at,  77 
Bernoulli's  theorem,  64 
Borda  mouthpiece,  61 
Branching  pipes,  88,  133 
Breast  wheel,  158 
Buoyancy,  20 

chamber,  Keokuk  lock,  39 


Canal  lock,  58 
Capacity  criterion,  184 
Catskill  aqueduct,  36,  140 

Venturi  meter,  69,  70 
Center  of  pressure,  13 
Centrifugal  pumps,  215 

characteristics,  227 

efficiency  and  design  of,  234 
Channel  cross  section,  103 
Characteristic  for  centrifugal  pumps, 
227 

speed,  185 
Characteristics    of    impulse    wheels 

and  turbines,  181 
Chezy's  formula,  85 
Church's  formula  for  water  hammer, 
204 


Circles,    circumferences    and    areas 

of,  268 
Circular    orifice,    efflux    coefficients 

for,  273 
sections,    hydraulic     properties 

of,  106 

Cock  in  circular  pipe,  80 
Complete  contraction,  49 
Compound  pipes,  86 
Conduits,  flow  in  open,  100 
Conical  mouthpiece,  63 
Contracted  weir,  51 
Contraction  coefficient,  45 
of  jet,  48 
of  section,  79 
effect  of,  92 

partial  and  complete,  49 
Crane,  hydraulic,  11 
Critical  velocity  of  water  in  pipes, 

71 

Cross  section  of  channel,  103 
Current  meter,  108 
wheel,  156,  158 

D 

Darcy's     modification     of     Pitot's 
tube,  110 

Deflection  of  jet,  147 

Density  of  water,  3 

Design  of  centrifugal  pumps,  234 

Diffusion  vanes,  221 

Diffusor,  pressure  developed  in,  226 

Discharge,  conditions  for  maximum, 

103 

equivalents,  260 
from  rectangular  orifice,  47 
of  rectangular  notch  weir,  48 
through  sharp-edged  orifice,  48 

Divided  flow,  86 

Doble  bucket,  160 

Draft  tube,  theory  of,  173 

Dry  dock,  floating,  236 

DuBuat's  paradox,  117 


291 


292 


INDEX 


Dynamic  pressure,  145 

in  bends  and  elbows,  149 

E 

Effective  fire  stream,  height  of,  90 

head,  46 
Efficiency  curves  for  turbines,  249, 

250    . 

of  hydraulic  press,  7 
maximum,  for  vanes,  151 
Efflux  coefficients,  46 

for  circular  orifice,  273 
for  square  orifice,  274 
Elasticity,    bulk    modulus    of,    for 

water,  202 
of  water,  1 

Elevator,  hydraulic,  12 
Empirical  weir  formula,  51,  52 
Enlargement  of  section,  78 

effect  of,  92 
Equilibrium  of  floating  bodies,   20 

of  fluids  in  contact,  16 
Experiments  on  flow  of  water,  90 

F 

Fire  nozzles,  64 

pumps,  239 

streams,  89 

stream  table,  275 
Flat  plate  in  current,  93 
Float  measurements,  108 
Floating  equilibrium,  21 

gate,  Keokuk  lock,  40 
Flow  in  small  pipes,  velocity  of,  74 
Force  pump,  209 
Fourneyron  type  of  turbine,  169 
Francis  type  of  turbine,  170 
Freeman's      experiments      on      fire 

streams,  89 

Friction  head  in  pipes,  278 
loss  in  pipe  flow,  77 


Gate  valve  in  circular  pipe,  80 
Gibson's     experiments     on     water 

hammer,  204 

Girard  impulse  turbine,  164 
Gradient,  hydraulic,  82 


H 

Head  and  pressure  equivalents,  47, 
259 

developed  by  centrifugal  pump, 
231 

effective,  on  orifice,  46 

lost  in  pipe  flow,  76 
Hele-Shaw's  experiments,  92 
Hook  gage,  54 
House  service  pipes,  129 
Hydraulic  accumulator,  9 

crane,  11 

dredge,  239 

efficiency  of  centrifugal  pumps, 
235 

elevator,  12 

gradient,  82 
slope  of,  83 

intensifier,  8 

jack,  10 

mining,  242 

motors,  types  of,  156 

press,  5 

radius,  84 
Hydrostatic  pressure,  3 


Ideal  velocity  head,  44 
Impact  on  plane  surface,  144 

on  surface  of  revolution,  146 

tube,  theory  of,  113 
Impeller,  pressure  developed  in,  224 
Impellers  for  centrifugal  pumps,  218 
Impulse  wheel,  152,  156 

wheels,  classification  of,  189 
Intensifier,  hydraulic,  8 

J 

Jack,  hydraulic,  10 
Jonval  type  of  turbine,  169 
Joukovsky's  experiments  on  water 
hammer,  204 

K 

Kensico  dam,  35,  36 
Keokuk  turbines,  179 

power  plant,  37 
Kinetic  pressure,  64 

head,  66,  82 


INDEX 


293 


Ku tier's  coefficients,  283 
formula,  101,  102 


Liquid  vein,  44 
Lock,  Keokuk,  38 
Loss  of  head  in  small  pipes,  74 
Lost  head  in  pipe  flow,  76 
summary  of,  81 

M 

Marietta's  flask,  60 
Mechanics,  principles  of,  288 
Mercury  barometer,  18 

pressure  gage,  19 
Metacenter,  23 

coordinates  of,  25 
Metacentric  height,  25 
Mine  drainage,  237 
Mississippi  River  Power  Co.,  37,  179 
Modulus  of  elasticity  for  water,  202 
Motors,  classification  of  hydraulic, 

157 
Moving  vanes,  149 

N 

Natural  channels,  flow  in,  106 
New  York  State  barge  canal,  96 
Niagara  Falls  power  plant,  172 
Non-sinuous  flow,  72 
Nozzle  diameter,  determination  of, 

198 

for  maximum  power,  196 
Nozzles,  flow  through,  60 

O 

Ontario  Power  Co.,  172 

Operating  range,  normal,  190 

Oval  sections,  hydraulic  properties 

of,  106 
Overshot  wheel,  159 


Packing,  frictional  resistance  of,  7 
Parallel  flow,  72 
Partial  contraction,  49 
Pelton  wheel,  159 

efficiency  of,  162 
Period  of  oscillation  of  ship,  27 


Piezometer,  18 

Pipe  flow,  ordinary,  75 

friction,  coefficient  of,  277 

lines,      power      transmitted 

through,  195 
Pipes,  dimensions  of,  265 

friction  head  in,  278 
Pitotmeter,  112 
Pitot  recorder,  113 

tube,  109,  116 
Poncelet  wheel,  158 
Power  and  efficiency,   relation  be- 
tween, 197 
Pressure  gage,  mercury,   19 

hydraulic,  5 

machines,  8 

static  and  dynamic,  145 
Properties  of  water,  260 
Pump  sizes,  calculation  of,  211 
Pumps,    capacity   of   reciprocating, 
266 

centrifugal,  215 

displacement,  207 

steam,  210 

suction,  207 

R 

Ram,  efficiency  of  hydraulic,  206 

hydraulic,  205 
Rate  of  flow  controller,  70 
Rational  weir  formulas,  51 
Reaction  of  jet,  154 
Reaction  turbine,  principle  of  oper- 
ation, 155,  157 

turbines,  169 

classification  of,  188 
Rectangular  notch  weir,  51 

discharge  from,  48 
weirs,  coefficients  for,  285-287 

orifice,  discharge  from,  47 
Relative  velocity  of  jet  and  vane,  149 
Reversal  of  jet,  147 
Rolling  and  pitching  of  ships,  27 
Rotation  of  liquids,  198 

S 

Siphon  lock,  100 
spillway,  96 
steel  pipe,  142 


294 


INDEX 


Siphons,  95 
Specific  discharge,  186 
power,  187 
speed,  186,  187 
weight,  3 

determination  of,  22 
physical  definition  of,  21 
weights  of  various   substances, 

264 

Speed  criterion,  182 
Square  orifice,  efflux  coefficients  for, 

274 

Stage  pumps,  223 
Standard  mouthpiece,  60 
Steam  pump,  210 
Stock  runner,  selection  of,  191 
Stream  gaging,  106 
line,  44 

motion,  93 
mouthpiece,  61 

Submerged  surfaces,  pressure  on,  12 
Suction  lift,  208 
Suppressed  weir,  51 

weirs,  discharge  coefficients  for, 

290 
Surface  of  liquid  in  rotation,  199 


Tanks,  connected,  rise  and  fall  in,  58 

filling  and  emptying,  56 
Throttle  valve  in  circular  pipe,  81 
Torricelli's  theorem,  45 
Translation,  effect  of,  198 
Tubes,  short,  flow  through,  60 


Turbine  pumps,  223 
setting,  177 
test  data,  251-255 

U 
Undershot  wheel,  158 


Vanes,  pressure  of  jet  on,  149 
Varying  head,  58 
Velocity  head,  ideal,  44 

of  approach,  49 

of  flow,  actual,  45 

variation  with  depth,  108 
Venturi  adjutage,  63 

meter,  67 

Catskill  aqueduct,  69,  70 
Viscosity,  2 

coefficient,  71 
Volute  casing,  219 
Vortex  chamber,  219 

W 

Water  barometer,  17 

hammer  in  pipes,  202 
properties  of,  260 

Waves  in  pipes,  pressure,  203 
period  of  compression,  203 
velocity  of  compression,  203 

Weights  and  measures,  263 

Weir  formulas,  51,  52 
measurements,  53 

Weirs,  construction  of,  53 
proportioning,  55 


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a*  r*  **'          1   '*» 

-    -  .. 

FEB    19  1935 

FEB   24W37 

,r    ^ 

12 

1 

' 
0        '&<*. 

\lf\,  . 

***  181940 

LD  21-100m-8,'34 

^X>0 


X 


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